Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{ll}\\ 11.&\textrm{Perhatikanlah pernyataan-pernyataan berikut}\\ &(1)\quad \displaystyle \sum_{i=1}^{5}(5i+2)=4\displaystyle \sum_{i=1}^{5}i+10\\ &(2)\quad \displaystyle \sum_{i=1}^{5}(5i^{2}-i)=5\displaystyle \sum_{i=1}^{5}i^{2}-\sum_{i=1}^{5}i\\ &(3)\quad \displaystyle \sum_{i=1}^{5}(3i-4)=3\displaystyle \sum_{i=1}^{5}i^{2}-4\\ &(4)\quad \displaystyle \sum_{i=1}^{5}(i+7i^{2})=\displaystyle \sum_{i=1}^{5}i-7\sum_{i=1}^{5}i\\ &\textrm{Pernyataan yang tepat ditunjukkan oleh}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&(1)\: \: \textrm{dan}\: \: (2)\\ \textrm{b}.&(1)\: \: \textrm{dan}\: \: (3)\\ \textrm{c}.&(1)\: \: \textrm{dan}\: \: (4)\\ \textrm{d}.&(2)\: \: \textrm{dan}\: \: (3)\\ \textrm{e}.&(2)\: \: \textrm{dan}\: \: (4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}(1)\quad&\displaystyle \sum_{i=1}^{5}(5i+2)=4\displaystyle \sum_{i=1}^{5}i+\sum_{i=1}^{5}2\\ &=4\displaystyle \sum_{i=1}^{5}i+5\times 2\\ &=4\displaystyle \sum_{i=1}^{5}i+10\\ (2)\quad&\displaystyle \sum_{i=1}^{5}(5i^{2}-i)=5\displaystyle \sum_{i=1}^{5}i^{2}-\sum_{i=1}^{5}i\\ (3)\quad&\displaystyle \sum_{i=1}^{5}(3i-4)=3\displaystyle \sum_{i=1}^{5}i^{2}-\sum_{i=1}^{5}4\\ &=3\displaystyle \sum_{i=1}^{5}i^{2}-5\times 4\\ &=3\displaystyle \sum_{i=1}^{5}i^{2}-20\\ (4)\quad&\displaystyle \sum_{i=1}^{5}(i+7i^{2})=\color{red}\displaystyle \sum_{i=1}^{5}i+7\sum_{i=1}^{5}i \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 12.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=1}^{4}i^{2}+\sum_{i=5}^{6}i^{2}\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 86\\ \color{red}\textrm{b}.&\displaystyle 91\\ \textrm{c}.&\displaystyle 95\\ \textrm{d}.&\displaystyle 101\\ \textrm{e}.&\displaystyle 105 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \sum_{i=1}^{4}i^{2}+\sum_{i=5}^{6}i^{2}&=\displaystyle \sum_{i=1}^{6}i^{2}\\ &=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}\\ &=1+4+9+16+25+36\\ &=\color{red}91 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=2}^{5}\left ( 4i^{2}-2i \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 144\\ \textrm{b}.&\displaystyle 148\\ \textrm{c}.&\displaystyle 154\\ \textrm{d}.&\displaystyle 164\\ \color{red}\textrm{e}.&\displaystyle 188 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\displaystyle \sum_{i=2}^{5}\left ( 4i^{2}-2i \right )\\ &=\left ( 4.2^{2}-2.2 \right )+\left ( 4.3^{2}-2.3 \right )+\left ( 4.4^{2}-2.4 \right )+\left ( 4.5^{2}-2.5 \right )\\ &=12+30+56+90\\ &=\color{red}188 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\textrm{Bentuk}\: \: 11^{n}-1\: \: \textrm{dengan}\: \: n\: \: \textrm{bilangan asli}\\ &\textrm{akan habis dibagi oleh}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 7\\ \textrm{b}.&\displaystyle 9\\ \color{red}\textrm{c}.&\displaystyle 10\\ \textrm{d}.&\displaystyle 11\\ \textrm{e}.&\displaystyle 13 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Bentuk}&\: \: 11^{n}-1\\ \textrm{untuk}&\: \: n=1\\ &=11^{1}-1\\ &=\color{red}10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Rumus yang tepat untuk pola}\: \: 12,13,14,15,...\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle U_{n}=n+9\\ \textrm{b}.&\displaystyle U_{n}=n+10\\ \color{red}\textrm{c}.&\displaystyle U_{n}=n+11\\ \textrm{d}.&\displaystyle U_{n}=2n+10\\ \textrm{e}.&\displaystyle U_{n}=2n+11 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\textrm{Bentuk}&\: \: 12,13,14,15,...\\ \textrm{untuk}&\: \: U_{n}=pn+q\\ 12&=p+q\\ 13&=2p+q\\ \textrm{akan}&\: \textrm{didapatkan}\\ &\begin{cases} p & =1 \\ q & =11 \end{cases}\\ \textrm{Sehing}&\textrm{ga}\\ U_{n}&=\color{red}n+11 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 16.&\textrm{Diketahui}\: \: 1+5+9+...+(4n-1)=2n^{2}-n\\ &\textrm{dengan}\: \: n\: \: \textrm{bilangan asli}.\: \textrm{Jika}\: \: m<k\: \: \textrm{dengan}\\ &m,k\: \: \textrm{bilangan asli juga},\: \textrm{maka}\\ &(4m-3)+(4m+1)+...+(4k-3)=....\\ &\begin{array}{llll}\\ \textrm{a}.&(k-m)(2k+2m-2)\\ \color{red}\textrm{b}.&(k-m+1)(2k+2m-3)\\ \textrm{c}.&(k-m+1)(2k-2m+1)\\ \textrm{d}.&(k-m+1)(2k^{2}+2m^{2}-3)\\ \textrm{e}.&(k-m)^{2}(2k-2m+4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&1+5+9+...+(4m-3)+(4m+1)+...+(4k-3)\\ &=\underset{2k^{2}-k}{\underbrace{1+5+...+(4k-3)}}-\underset{2(m-1)^{2}-(m-1)}{\underbrace{1+5+...+(4(m-1)-3)}}\\ &=2k^{2}-k-\left ( 2(m-1)^{2}-(m-1) \right )\\ &=2k^{2}-k-2(m-1)^{2}+(m-1)\\ &=2k^{2}-k-2\left ( m^{2}-2m+1 \right )+m-1\\ &=2k^{2}-k-2m^{2}+4m-2+m-1\\ &=2k^{2}-k-2m^{2}+5m-3\\ &=\color{red}(k-m+1)(2k+2m-3) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 17.&\textrm{Diketahui}\: \: 2^{1}+2^{2}+2^{3}+...+2^{n}=2^{n+1}-2\\ &\textrm{dengan}\: \: n\: \: \textrm{bilangan asli}.\: \textrm{Jika}\: \: k\: \: \textrm{bilangan asli},\\ &\textrm{maka}\: \: 2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}=....\\ &\begin{array}{llll}\\ \textrm{a}.&(k-m)(2k+2m-2)\\ \textrm{b}.&(k-m+1)(2k+2m-3)\\ \textrm{c}.&(k-m+1)(2k-2m+1)\\ \color{red}\textrm{d}.&(k-m+1)(2k^{2}+2m^{2}-3)\\ \textrm{e}.&(k-m)^{2}(2k-2m+4) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}\\ &=2^{1}+2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}-2^{1}\\ &=\underset{2^{k+1+1}-2}{\underbrace{2^{1}+2^{2}+2^{3}+2^{4}+...+2^{k}+2^{k+1}}}-2^{1}\\ &=2^{k+2}-2-2\\ &=2^{k+2}-4\\ &=2^{k}.2^{2}-4\\ &=2^{k}\times 4-4\\ &=4\left ( 2^{k}-1 \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 18.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &2+5+10+17+...+\left ( n^{2}+1 \right )=\displaystyle \frac{1}{6}(n+1)\left ( 2n^{2}+n+6 \right )\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k,\: \: \textrm{maka}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&2+5+10+17+...+\left ( k^{2}+1 \right )\\ &=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{b}.&2+5+10+17+...+\left ( n^{2}+1 \right )\\ &=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{c}.&2+5+10+17+...+\left ( k^{2}+2 \right )\\ &=\displaystyle \frac{1}{6}(k+2)\left ( 2k^{2}+5k+9 \right )\\ \textrm{d}.&\left ( k^{2}+1 \right )=\displaystyle \frac{1}{6}(k+1)\left ( 2k^{2}+k+6 \right )\\ \textrm{e}.&\left ( n^{2}+2 \right )=\displaystyle \frac{1}{6}(n+1)\left ( 2n^{2}+5n+9 \right ) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Cukup jelas}\\ &\textrm{Tinggal mensubstitusikan dari}\\ &\textrm{tiap}\: \: \color{red}n\: \: \color{black}\textrm{diganti}\: \: \color{red}k \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &12+17+22+...+\left ( 5n+7 \right )=\displaystyle \frac{1}{2}(n+1)(5n+14)\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k,\: \: \textrm{maka benar}\\ &\textrm{untuk}\: \: n=k+1.\: \textrm{Pernyataan ini dapat}\\ &\textrm{dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+14)\\ \textrm{b}.&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+19)\\ \textrm{c}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+1)(5k+19)\\ \textrm{d}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+2)(5k+14)\\ \color{red}\textrm{e}.&12+17+22+...+\left ( 5k+12 \right )=\displaystyle \frac{1}{2}(k+2)(5k+19) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&12+17+22+...+\left ( 5k+7 \right )=\displaystyle \frac{1}{2}(k+1)(5k+14)\\ &12+17+22+...+\left ( 5(k+1)+7 \right )\\ &\qquad\qquad\qquad\quad=\displaystyle \frac{1}{2}((k+1)+1)(5(k+1)+14)\\ &12+17+22+...+\left ( 5k+12 \right )=\color{red}\displaystyle \frac{1}{2}(k+2)(5k+19) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 20.&\textrm{Diketahui bahwa}\: \: S(n)\: \: \textrm{adalah formula dari}\\ &4+5+6+7+...+(n+3)<5n^{2}\\ &\textrm{Jika}\: \: S(n)\: \: \textrm{benar, untuk}\: \: n=k+1,\: \: \textrm{maka}\\ &\textrm{pernyataan ini dapat ditulis dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&4+5+6+...+(k+4)<5k^{2}\\ \textrm{b}.&4+5+6+...+(k+3)<5k^{2}\\ \textrm{c}.&4+5+6+...+(k+3)<5(k+1)^{2}\\ \color{red}\textrm{d}.&4+5+6+...+(k+4)<5(k^{2}+2k+1)\\ \textrm{e}.&4+5+6+...+(k+4)<5(k+1)(k-1) \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&4+5+6+...+(n+3)<5n^{2}\\ &\textrm{Saat}\: \: n=k+1,\: \: \textrm{maka}\\ &4+5+6+...+((k+1)+3)<5(k+1)^{2}\\ &=4+5+6+...+(k+4)<\color{red}5\left ( k^{2}+2k+1 \right ) \end{aligned} \end{array}$.

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas XI

 $\begin{array}{ll}\\ 1.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=1}^{6}16i\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&306\\ \textrm{b}.&314\\ \textrm{c}.&326\\ \color{red}\textrm{d}.&336\\ \textrm{e}.&402 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \sum_{i=1}^{6}16i&=16.1+16.2+16.3+16.4+16.5+16.6\\ &=16+32+48+64+80+96\\ &=\color{red}336 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Hasil dari}\: \: \displaystyle \sum_{i=2}^{9}i^{2}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&274\\ \textrm{b}.&278\\ \textrm{c}.&280\\ \color{red}\textrm{d}.&284\\ \textrm{e}.&286 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\displaystyle \sum_{i=2}^{9}i^{2}&=2^{2}+3^{2}+4^{2}+5^{2}+..+9^{2}\\ &=4+9+16+25+...+81\\ &=\color{red}284 \end{aligned}\end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Poa bilangan}\: \: 12,14,16,18,20,...,(2n+10).\\ &\textrm{Nilai suku ke-100 adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.&180\\ \textrm{b}.&194\\ \textrm{c}.&198\\ \textrm{d}.&208\\ \color{red}\textrm{e}.&210\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}U_{n}&=2n+10\\ U_{100}&=2\times 100+10\\ &=210 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa jika}\\ & 31+39+47+\cdots +8n+23=4n^{2}+27n\\ & \textrm{dengan}\: \: k,n\in \mathbb{N}\: \: \textrm{maka}\\ & 31+39+47+\cdots +8n+23+8k+31=....\\ &\begin{array}{lllllll}\\ \textrm{a}.&4k^{2}+27k\\ \textrm{b}.&4k^{2}+35k\\ \color{red}\textrm{c}.&4k^{2}+35k+31\\ \textrm{d}.&4k^{2}+35k+1\\ \textrm{e}.&4k^{2}+35k+54\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\underset{4k^{2}+27k}{\underbrace{31+39+47+\cdots +8k+23}}+8k+31\\ &=4k^{2}+27k+8k+31\\ &=\color{red}4k^{2}+35k+31 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Dengan Induksi Matematika untuk}\: \: n\in \mathbb{N}\\ &\textrm{dapat dibuktikan bahwa}\: \: n(n+1)(n+2)\\ &\textrm{akan habis dibagi oleh}\\ &\begin{array}{lllllll}\\ \textrm{a}.&4\\ \textrm{b}.&5\\ \color{red}\textrm{c}.&6\\ \textrm{d}.&7\\ \textrm{e}.&8\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}P(n)&=n(n+1)(n+2)\\ P(1)&=1(1+1)(1+2)=\color{red}1.2.3\\ &\color{purple}\textrm{adalah bilangan yang habis dibagi 6} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Dengan Induksi Matematika untuk}\: \: n\in \mathbb{N}\\ &\textrm{dapat dibuktikan juga bahwa}\: \: 7^{n}-2^{n}\\ &\textrm{akan habis dibagi oleh}\\ &\begin{array}{lllllll}\\ \textrm{a}.&2\\ \textrm{b}.&3\\ \textrm{c}.&4\\ \color{red}\textrm{d}.&5\\ \textrm{e}.&6\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}P(n)&=7^{n}-2^{n}\\ P(1)&=7^{1}-2^{1}\\ &=7-2=\color{red}5\\ &\textrm{adalah bilangan yang habis dibagi 5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Diketahui bahwa}\: \: P(n)\: \: \textrm{rumus dari}\\ & 3+6+9+\cdots +3n=\displaystyle \frac{3}{2}n(n+1)\\ &\textrm{maka langkah pertama dengan induksi matematika}\\ &\textrm{dalam pembuktian rumus tersebut adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.&P(n)\: \: \textrm{benar untuk}\: \: n=-1\\ \color{red}\textrm{b}.&P(n)\: \: \textrm{benar untuk}\: \: n=1\\ \textrm{c}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan bulat}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan rasional}\\ \textrm{d}.&P(n)\: \: \textrm{benar untuk}\: \: n\: \: \textrm{bilangan real} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\textrm{Langkah}&\: \textrm{awal yang harus ditunjukkan adalah}\\ n=1&\: \: \textrm{atau}\: \: P(1)\: \: \textrm{harus benar, yaitu}:\\ P(1)&=3.1(\textit{ruas kiri})=\displaystyle \frac{3}{2}.1.(1+1)(\textit{ruas kanan})=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 8.&\textrm{Bila kita hendak membuktikan}\: \: \displaystyle \sum_{i=1}^{n}=\displaystyle \frac{1}{2}n(n+1)\\ &\textrm{dengan induksi matematika}\\ &\textrm{maka untuk langkah}\: \: n=k+1\\ &\textrm{bentuk yang harus ditunjukkan adalah}...\\ &\begin{array}{lll}\\ \textrm{a}.&1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ \textrm{b}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ \textrm{c}.&1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+2)\\ \color{red}\textrm{d}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+1)(k+2)\\ \textrm{e}.&1+2+3+\cdots +k+(k+1)=\displaystyle \frac{1}{2}(k+2)(k+3)\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}P(n)&=1+2+3+\cdots +n=\displaystyle \frac{1}{2}n(n+1)\\ P(k)&=1+2+3+\cdots +k=\displaystyle \frac{1}{2}k(k+1)\\ P(k+1)&=1+2+3+\cdots +k+(k+1)\\ &=\underset{\displaystyle \frac{1}{2}k(k+1)}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ &=\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left ( \displaystyle \frac{1}{2}k+1 \right )\\ &=(k+1)\displaystyle \frac{1}{2}(k+2)\\ &=\color{red}\displaystyle \frac{1}{2}(k+1)(k+2) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n-1}{n+3},\: \textrm{maka}\: \: P(k+1)\\ & \textrm{dinyatakan dengan}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k-1}{k+3}\\ \textrm{b}.&\displaystyle \frac{k-1}{k+4}\\ \color{red}\textrm{c}.&\displaystyle \frac{k}{k+4}\\ \textrm{d}.&\displaystyle \frac{k+1}{k+4}\\ \textrm{e}.&\displaystyle \frac{k+1}{k+5} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}P(n)=&\displaystyle \frac{n-1}{n+3}\\ P(k+1)&=\displaystyle \frac{k+1-1}{k+1+3}\\ &=\color{red}\displaystyle \frac{k}{k+4} \end{aligned}\end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: P(n)=\displaystyle \frac{n^{2}+1}{4},\: \: \textrm{maka}\\ &\textrm{pernyataan untuk}\: \: P(k+1)\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle \frac{k^{2}+2k+1}{4}\\ \color{red}\textrm{b}.&\displaystyle \frac{k^{2}+2k+2}{4}\\ \textrm{c}.&\displaystyle \frac{k^{2}+2k+2}{5}\\ \textrm{d}.&\displaystyle \frac{k^{2}+2k+3}{5}\\ \textrm{e}.&\displaystyle \frac{k^{2}+2k+3}{4} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}P(n)&=\displaystyle \frac{n^{2}+1}{4}\\ P(k+1)&=\displaystyle \frac{(k+1)^{2}+1}{4}\\ &=\color{red}\displaystyle \frac{k^{2}+2k+2}{4} \end{aligned} \end{array}$

Latihan Soal 11 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 101.&\textrm{Nilai dari} \: \sin 1020^{\circ}=....\\ &\textrm{a}.\quad -1\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{1}{2}\sqrt{3}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{3}\\\\ &\textrm{Jawab}:\qquad\color{red}\textbf{b}\\ &\begin{aligned}\sin 1020^{\circ}&=\sin \left ( 3\times 360^{\circ}-60^{\circ} \right )\\ &=\sin \left ( 0^{\circ}-60^{\circ} \right )\\ &=\sin \left ( -60^{\circ} \right )\\ &=-\sin 60^{\circ}\\ &=\color{red}-\displaystyle \frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 102.&\textrm{Nilai dari} \: \cot (-1290)^{\circ}=....\\ &\textrm{a}.\quad \color{red}-\sqrt{3}\\ &\textrm{b}.\quad \displaystyle -\frac{1}{3}\sqrt{3}\\ &\textrm{c}.\quad \displaystyle \frac{1}{3}\sqrt{3}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{a}\\ &\begin{aligned}\cot (-1290)^{\circ}&=-\cot \left ( 3\times 360^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 0^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 210^{\circ} \right )\\ &=-\frac{1}{\tan 210^{\circ}}\\ &=-\frac{1}{\tan \left ( 180^{\circ}+30^{\circ} \right )}\\ &=-\frac{1}{\tan 30^{\circ}}\\ &=\color{red}-\displaystyle \sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 103.&\textrm{Nilai dari}\\ & \sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\sqrt{3}\qquad&&\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{b}.\quad \color{red}\displaystyle -\frac{1}{2}\sqrt{3}&\qquad&\textrm{e}.\quad \displaystyle \frac{1}{3}\sqrt{3}\\ \textrm{c}.\quad \displaystyle -\frac{1}{2} \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{b}\\ &\begin{aligned}&\sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}\\ &=\sin \left ( 180^{\circ}+60^{\circ} \right )+\sin \left ( 180^{\circ}+45^{\circ} \right )+\cos \left ( 360^{\circ}-45^{\circ} \right )\\ &=-\sin 60^{\circ}+\left [ -\sin 45^{\circ} \right ]+\cos 45^{\circ}\\ &=\left ( -\frac{1}{2}\sqrt{3} \right )+\left ( -\frac{1}{2}\sqrt{2} \right )+\frac{1}{2}\sqrt{2}\\ &=\color{red}-\frac{1}{2}\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 104.&\textrm{Nilai dari} \\ & \displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\\ \textrm{b}.\quad \displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}\\ \textrm{c}.\quad \displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}\\ \textrm{d}.\quad \color{red}\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}\\ \textrm{e}.\quad \displaystyle \frac{1+2\sqrt{3}}{1-2\sqrt{3}} \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{d}\\ &\begin{aligned}&\displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}\\ &=\displaystyle \frac{\sin 30^{\circ}+\sin \left ( 180^{\circ}-30^{\circ} \right )+\cos \left ( 360^{\circ}-30^{\circ} \right )}{\tan 45^{\circ}+\cos \left ( 180^{\circ}+30^{\circ} \right )}\\ &=\displaystyle \frac{\sin 30^{\circ}+\sin 30^{\circ}+\cos 30^{\circ}}{\tan 45^{\circ}-\cos 30^{\circ}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\color{red}\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 105.&\textrm{Nilai dari} \\ &\displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -2\qquad&&\textrm{d}.\quad 1\\ \textrm{b}.\quad \displaystyle -\frac{1}{2}&\textrm{c}.\quad \displaystyle \frac{1}{2}\qquad&\textrm{e}.\quad 2 \end{array}\\ &\textrm{Jawab}:\qquad\color{red}\textrm{e}\\ &\begin{aligned}&\displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}\\ &=\displaystyle \frac{\sin 270^{\circ}\times \cos \left ( 180^{\circ}-45^{\circ} \right )\times \tan \left ( 180^{\circ}-45^{\circ} \right )}{\sin \left ( 180^{\circ}-30^{\circ} \right )\times \cos \left ( 180^{\circ}+45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left (-\cos 45^{\circ} \right )\times \left ( - \tan 45^{\circ}\right )}{\sin 30^{\circ}\times \left ( -\cos 45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\times -1}{\displaystyle \frac{1}{2}\times \left ( -\frac{1}{2}\sqrt{2} \right )}\\ &=\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{-\frac{1}{4}\sqrt{2}}\\ &=\color{red}\displaystyle 2\end{aligned} \end{array}$

$\begin{array}{ll}\\ 106.&\textrm{Perhatikanlah gambar kurva berikut ini}\end{array}$.

$\begin{array}{ll}\\ .\, \: \: \quad&\begin{array}{llllll}\\ \textrm{a}.&y=-2\cos 2x\\ \textrm{b}.&y=2\cos \displaystyle \frac{3}{2}x\\ \textrm{c}.&y=-2\cos \displaystyle \frac{3}{2}x\\ \textrm{d}.&y=2\sin \displaystyle \frac{3}{2}x\\ \textrm{e}.&y=-2\sin \displaystyle \frac{3}{2}x\\\\ &(\textbf{SIMAK UI 2009 Mat Das}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&\textrm{Dari gambar tampak jelas bahwa}\\ &\textrm{grafik di atas atas adalah grafik}\: \textbf{fungsi cosinus}\\ &\textrm{dengan amplitudo 2 dan periodenya}\: :\: \displaystyle \frac{3}{2}\pi \\ &\textrm{Maka persamaan grafiknya adalah}:\\ &y=2\cos \displaystyle \frac{3}{2}\pi \\ &\textrm{Karena posisinya terbalik, maka}\\ &y=\color{red}-2\cos \displaystyle \frac{3}{2}\pi \\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 107.&\textrm{Nilai minimum jika}\\ & f(x)=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\\ & \textrm{adalah}=....\\ &\begin{array}{llllll}\\ \textrm{a}.&2005\\ \textrm{b}.&2007\\ \textrm{c}.&\color{red}2011\\ \textrm{d}.&2013\\ \textrm{e}.&\textrm{tidak ada satupun jawaban dari a sampai d}\\\\ &(\textbf{NUS Mathematics A Level}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&f(x)=\left ( 2004\cos 2005x-2006 \right )^{2}+2007\\ &\textrm{Supaya bernilai minimum, }\\ &\textrm{maka nilai}\quad \color{red}\cos 500x=1,\\ & \color{black}\textrm{ingat nilai}\: \: -1\leq \cos n\pi \leq 1\\ &\textrm{maka},\\ &f_{min}=\left ( 2004.1-2006 \right )^{2}+2007\\ &=(-2)^{2}+2007\\ &=4+2007\\ &=\color{red}2011 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 108.&\textrm{Penyelesaian persamaan}\\ & \cos ^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \textrm{adalah}\, ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\pi -\cot ^{-1}\left ( \displaystyle \frac{1}{2} \right )\\ \textrm{b}.&\pi +\tan ^{-1}\left ( \displaystyle \frac{1}{2} \right )\\ \textrm{c}.&\pi -\cot ^{-1}\left ( -1 \right )\\ \textrm{d}.&\color{red}\pi +\tan ^{-1}\left ( -\displaystyle \frac{1}{2} \right )\\ \textrm{e}.&\pi -\tan ^{-1}\left ( \displaystyle \frac{1}{4} \right ) \\ \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Perhatikan bahwa},\\ &\cos ^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ &\cos x\left ( \cos x-2 \right )=2\sin x\left ( 2-\cos x \right )\\ &\cos x\left ( \cos x-2 \right )=-2\sin x\left ( \cos x -2\right )\\ &\left (\cos x+2\sin x \right )\left ( \cos x-2 \right )=0\\ &\left (\cos x+2\sin x \right )=0\: \: \textrm{atau} \: \: \left ( \cos x-2 \right )=0\\ &2\sin x=-\cos x\: \: \textbf{(mm)}\: \: \textrm{atau}\: \: \cos x=2\: \: \textbf{(tm)}\\ &\textrm{maka}\\ &\displaystyle \frac{\sin x}{\cos x}=-\displaystyle \frac{1}{2}\: \: \Leftrightarrow \: \: \tan x=-\displaystyle \frac{1}{2}\\ &x=\color{red}\tan ^{-1}\left ( -\displaystyle \frac{1}{2} \right )+k.\pi ,\quad k\in \mathbb{Z} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 109.&\textrm{Jika diketahui bahwa}\\ & \sin \beta -\tan \beta -2\cos \beta +2=0\\ & \textrm{dengan}\: \: 0<\beta <\displaystyle \frac{\pi }{2},\\ &\textrm{maka himpunan harga}\: \: \sin \beta =....\\ &\begin{array}{llllll}\\ \textrm{a}.&\color{red}\left \{ \displaystyle \frac{2}{5}\sqrt{5} \right \}\\ \textrm{b}.&\left \{ 0 \right \}\\ \textrm{c}.&\left \{ \displaystyle \frac{2}{5}\sqrt{5},0 \right \}\\ \textrm{d}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5} \right \}\\ \textrm{e}.&\left \{ \displaystyle \frac{1}{5}\sqrt{5},0 \right \}\\\\ &\textbf{(SIMAK UI 2009)} \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{a}\\ &\begin{aligned}&\sin \beta -\tan \beta -2\cos \beta +2=0\\ &\sin \beta -\displaystyle \frac{\sin \beta }{\cos \beta } -2\cos \beta +2=0\\ &\sin \beta \cos \beta -\sin \beta -2\cos^{2} \beta +2\cos \beta =0\\ &\sin \beta \left ( \cos \beta -1 \right )-2\cos \beta \left ( \cos \beta -1 \right )=0\\ &\left ( \sin \beta -2\cos \beta \right )\left ( \cos \beta -1 \right )=0\\ &\left ( \sin \beta -2\cos \beta \right )=0\: \textbf{(mm)}\\ &\qquad \textrm{atau}\quad \left ( \cos \beta -1 \right )=0\: \textbf{(tmm)}\\ &\textrm{maka},\\ &\left ( \sin \beta -2\cos \beta \right )=0\\ &\sin \beta =2\cos \beta\\ &\displaystyle \frac{\sin \beta }{\cos \beta }=2\\ &\tan \beta =2=\displaystyle \frac{2}{1},\qquad \textrm{buatlah ilustrasi} \\ &\textrm{dengan membuat segitiga siku-siku}.\\ &\textrm{Sehingga akan didapatkan nilai}\\ &\sin \beta =\color{red}\displaystyle \frac{2}{5}\sqrt{5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 110.&\textrm{Diketahui bahwa}\\ &\sin \theta -\cos \theta =\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}\: \: \textrm{dan}\\ & \cos ^{3}\theta -\sin ^{3}\theta =\displaystyle \frac{1}{a}\left ( b\sqrt{5}-c\sqrt{3} \right ),\\ &\textrm{dengan}\: \: a,\: b,\: c\: \: \textrm{adalah bilangan asli, maka}\\ &(1) \quad b-c>0\\ &(2) \quad a-b=7\\ &(3)\quad a-3b+c=0\\ &(4)\quad a+b+c=12\\ &\begin{array}{llllll}\\ \textrm{a}.&(1),\: (2).\: \textrm{dan}\: (3)\: \textrm{benar}\\ \textrm{b}.&(1),\: \textrm{dan}\: (3)\: \textrm{benar}\\ \textrm{c}.&\color{red}(2),\: \textrm{dan}\: (4)\: \textrm{benar}\\ \textrm{d}.&\textrm{hanya}\: (4)\: \textrm{yang benar}\\ \textrm{e}.&\textrm{semuanya benar}\\\\ &(\textbf{SIMAK UI 2015 Mat IPA}) \end{array}\\ &\\ &\textrm{Jawab}:\quad\color{red}\textbf{c}\\ &\begin{aligned}&\sin \theta -\cos \theta =\displaystyle \frac{\sqrt{5}-\sqrt{3}}{2}\\ &\sin ^{2}\theta +\cos^{2}\theta -2\sin \theta \cos \theta =\displaystyle \frac{8-2\sqrt{5}}{4}\\ &1-2\sin \theta \cos \theta =\displaystyle \frac{8-2\sqrt{5}}{4}\\ &\sin \theta \cos \theta =\displaystyle \frac{\sqrt{5}-2}{4}\\ &\textrm{maka},\\ &\cos ^{3}\theta -\sin ^{3}\theta \\ &\qquad=\left ( \cos \theta -\sin \theta \right )\left ( \cos ^{2}\theta +\sin \theta \cos \theta +\sin ^{2}\theta \right )\\ &=\left ( \displaystyle \frac{\sqrt{3}-\sqrt{5}}{2} \right )\left ( 1+\displaystyle \frac{\sqrt{5}-2}{4} \right )\\ &=\displaystyle \frac{1}{8}\left ( \sqrt{3}-\sqrt{5} \right )\left ( 2+\sqrt{5} \right )\\ &=\displaystyle \frac{1}{8}\left ( 2\sqrt{3}+3\sqrt{5}-2\sqrt{5}-5\sqrt{3} \right )\\ &=\displaystyle \frac{1}{8}\left (\sqrt{5}-3\sqrt{3} \right )=\displaystyle \frac{1}{a}\left ( b\sqrt{5}-c\sqrt{3} \right ) \left\{\begin{matrix} a=8\\ b=1\\ c=3 \end{matrix}\right.\\ &\textrm{sehingga}\\ &a-b=8-1=\color{red}7\\ &a+b+c=8+1+3=\color{red}12 \end{aligned} \end{array}$

Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll}\\ 91.&\textrm{Bentuk sederhana dari}\\ &\quad\quad 4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}2+2\sin 2x &&&\textrm{d}.&2+2\sin x\\ \textrm{b}.&\displaystyle 2+\sin 2x&&&\textrm{e}.&2+\sin x\\ \textrm{c}.&\displaystyle 2\sin 2x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&4\sin \left ( \displaystyle \frac{1}{4}\pi +x \right )\cos \left ( \displaystyle \frac{1}{4}\pi -x \right )\\ &=2\left ( \sin \left ( \displaystyle \frac{1}{2}\pi \right )+\sin (2x) \right )\\ &=2\left (1+\sin 2x \right )\\ &=2+2\sin 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 92.&\textrm{Bentuk sederhana dari}\\ & 2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}1-\sin 2x&&\textrm{d}.\quad \cos 2x\\\\ \textrm{b}.\quad 1+\sin 2x&\textrm{c}.\quad 1-\cos 2x&\textrm{e}.\quad -\cos 2x\end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{a}\\ &\begin{aligned}&2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )\\ &=2\sin \left ( 135^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\sin \left ( 90^{\circ}+45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos \left ( 45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos ^{2}\left ( 45^{\circ}+x \right )\\ &=2\left ( \cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x \right )^{2}\\ &=2\left ( \frac{1}{2}\sqrt{2}.\cos x-\displaystyle \frac{1}{2}\sqrt{2}.\sin x \right )^{2}\\ &=\displaystyle \left ( \cos x-\sin x \right )^{2}\\ &=\cos ^{2}x+\sin ^{2}x-2\sin x\cos x\\ &=\color{red}1-\sin 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 93.&\textrm{Bentuk sederhana dari}\\ & 2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos (2x+\pi )&&\textrm{d}.\quad -\cos \left ( 2x-\displaystyle \frac{\pi }{2} \right )\\ \textrm{b}.\quad \cos \left ( 2x+\displaystyle \frac{\pi }{2} \right )&&\textrm{e}.\quad \color{red}\sin 2x-1\\ \textrm{c}.\quad \cos 2x \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )\\ &=\cos \left ( x+\displaystyle \frac{\pi }{4}+\displaystyle \frac{3}{4}\pi -x \right )+\cos \left (x+\displaystyle \frac{\pi }{4} -\left ( \displaystyle \frac{3}{4}\pi -x \right ) \right )\\ &=\cos \pi +\cos \left ( 2x-\frac{2}{4}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2x \right ),\\ &\quad\quad \color{blue}\textrm{ingat bahwa}\: \cos (-\alpha )=\cos \alpha \\ &=-1+\sin 2x\\ &=\color{red}\sin 2x-1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 94.&\textrm{Nilai dari}\\ &\quad\quad \sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\displaystyle \frac{3}{8} &&&\textrm{d}.&\displaystyle \color{red}\frac{3}{8}\\\\ \textrm{b}.&\displaystyle -\frac{1}{8}&&&\textrm{e}.&\displaystyle \frac{5}{8}\\ \textrm{c}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sqrt{3}\sin 80^{\circ}\sin 160^{\circ}\sin 320^{\circ}\\ &=\sqrt{3}\sin 80^{\circ}\sin 20^{\circ}\left (-\sin 40^{\circ} \right )\\ &=-\sqrt{3}\sin 80^{\circ}\sin 40^{\circ}\sin 20^{\circ}\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{2}\left ( \cos 60^{\circ}-\cos 20^{\circ} \right ) \right )\\ &=-\sqrt{3}\sin 80^{\circ}\left ( -\displaystyle \frac{1}{4}+\displaystyle \frac{\cos 20^{\circ}}{2} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{2}\sqrt{3}\sin 80^{\circ}\cos 20^{\circ}\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 100^{\circ}+\sin 60^{\circ} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\left ( \sin 80^{\circ}+\displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}-\displaystyle \frac{1}{4}\sqrt{3}\sin 80^{\circ}+\displaystyle \frac{1}{8}\sqrt{9}\\ &=\displaystyle \frac{3}{8} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 95.&\textrm{Nilai dari}\\ &\quad\quad \cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{8} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle -\frac{1}{4}&\quad \textrm{c}.&0\quad &\textrm{e}.&\displaystyle \frac{1}{3} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textbf{Alternatif 1}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\left (\sin \displaystyle \frac{4\pi }{7}-\sin 0 \right )\frac{\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{4\pi }{7}\displaystyle \cos \frac{\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{2\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\left ( \sin \displaystyle \frac{5\pi }{7}+\sin \displaystyle \frac{3\pi }{7} \right )\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{5\pi }{7}\cos \displaystyle \frac{4\pi }{7}+\sin \displaystyle \frac{3\pi }{7}\cos \displaystyle \frac{4\pi }{7}}{4\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin \displaystyle \frac{9\pi }{7}+\sin \displaystyle \frac{\pi }{7}+\sin \displaystyle \frac{7\pi }{7}+\sin \left (-\displaystyle \frac{\pi }{7} \right )}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}+\sin \displaystyle \frac{\pi }{7}+0-\sin \displaystyle \frac{\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{-\sin \displaystyle \frac{2\pi }{7}}{8\sin \displaystyle \frac{2\pi }{7}}\\ &=-\displaystyle \frac{1}{8} \end{aligned}\\ &\textbf{Alternatif 2}\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\\ &=\cos \displaystyle \frac{4\pi }{7}\cos \frac{2\pi }{7}\cos \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{6\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( \cos \left ( \pi -\displaystyle \frac{\pi }{7} \right )+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &=\displaystyle \frac{1}{2}\left ( -\cos \displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7} \right )\cos \displaystyle \frac{\pi }{7}\\ &= \displaystyle \frac{1}{2}\left (-\cos ^{2}\displaystyle \frac{\pi }{7}+\cos \displaystyle \frac{2\pi }{7}\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos \displaystyle \frac{2\pi }{7}-\cos 0+\cos \displaystyle \frac{3\pi }{7}+\cos \displaystyle \frac{\pi }{7} \right )\\ &=\displaystyle \frac{1}{4}\left ( -\cos 0+\color{red}\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7} \color{black}\right )\\ &=\displaystyle \frac{1}{4}\left ( -1+\color{red}\displaystyle \frac{1}{2}\color{black} \right )\\ &=\displaystyle \frac{1}{4}\times \left (-\frac{1}{2} \right )\\ &=-\displaystyle \frac{1}{8} \end{aligned} \end{array}$.

Berikut penjelasan untuk  $\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}=\color{red}\displaystyle \frac{1}{2}$.

$\begin{aligned}&\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\\ &=\cos \displaystyle \frac{\pi }{7}-\cos \displaystyle \frac{2\pi }{7}+\cos \displaystyle \frac{3\pi }{7}\times \displaystyle \frac{\left (2\sin\displaystyle \frac{2\pi }{7} \right ) }{\left (2\sin\displaystyle \frac{2\pi }{7} \right )}\\ &=\displaystyle \frac{2\cos\displaystyle \frac{\pi }{7}\sin\displaystyle \frac{2\pi }{7}-2\cos\displaystyle \frac{2\pi }{7}\sin\displaystyle \frac{2\pi }{7}+2\cos\displaystyle \frac{3\pi }{7}\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\left (-\displaystyle \frac{\pi }{7} \right )-\left ( \sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{0\pi }{7} \right )+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}+\sin\displaystyle \frac{\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}-\sin\displaystyle \frac{\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{3\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{5\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\left (\pi -\displaystyle \frac{4\pi }{7} \right )-\sin\displaystyle \frac{4\pi }{7}+\sin\left (\pi -\displaystyle \frac{2\pi }{7} \right )}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{4\pi }{7}-\sin\displaystyle \frac{4\pi }{7}+\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{\sin\displaystyle \frac{2\pi }{7}}{2\sin\displaystyle \frac{2\pi }{7}}\\ &=\displaystyle \frac{1}{2}\qquad \blacksquare \end{aligned}$.

$\begin{array}{ll}\\ 96.&\textrm{Nilai dari}\\ &\quad\quad \sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{2}\\\\ \textrm{b}.&\displaystyle \color{red}\frac{1}{8}&\quad \textrm{c}.&\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle 1 \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhat}&\textrm{ikan bahwa}\\ \sin \displaystyle \displaystyle \frac{\pi }{14}&=\sin \left (\displaystyle \frac{7\pi }{14}-\frac{6\pi }{14} \right )=\sin \left ( \displaystyle \frac{1}{2}\pi -\frac{6\pi }{14} \right )\\ &=\cos \displaystyle \frac{6\pi }{14} \\ \sin \displaystyle \frac{3\pi }{14}&=...=\cos \displaystyle \frac{4\pi }{14}\\ \sin \displaystyle \frac{9\pi }{14}&=...=\sin \displaystyle \frac{5\pi }{14}=\cos \displaystyle \frac{2\pi }{14} \end{aligned}\\ &...\\ &\sin \displaystyle \frac{\pi }{14}\sin \frac{3\pi }{14}\sin \frac{9\pi }{14}\\ &=\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\cos \displaystyle \frac{2\pi }{14}\times \displaystyle \frac{2\sin \displaystyle \frac{2\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{6\pi }{14}\cos \displaystyle \frac{4\pi }{14}\sin \displaystyle \frac{4\pi }{14}}{2\sin \displaystyle \frac{2\pi }{14}}\\ &\textrm{silahkan dilanjutkan}\\ &...\\ &=\displaystyle \frac{1}{8} \end{array}$.

$\begin{array}{ll}\\ 97.&\textrm{Nilai dari}\\ & \cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}-\displaystyle \frac{1}{16} &&&\textrm{d}.&\displaystyle \frac{1}{16}\\\\ \textrm{b}.&\displaystyle \frac{1}{8}&\quad \textrm{c}.&\displaystyle 0\quad &\textrm{e}.&\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{8\pi }{5}\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \left (\pi +\displaystyle \frac{3\pi }{5} \right )\\ &=\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\left (-\cos \displaystyle \frac{3\pi }{5} \right )\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{4\pi }{5}\cos \displaystyle \frac{3\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\\ &=-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\cos \displaystyle \frac{4\pi }{5}\times \displaystyle \frac{2\sin \displaystyle \frac{\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\frac{-\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5}\left ( \sin \pi -\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \frac{2\pi }{5}\cos \frac{3\pi }{5} \sin \frac{3\pi }{5}}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \cos \displaystyle \frac{2\pi }{5}\sin \displaystyle \frac{3\pi }{5} \right )}{2\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\left ( \sin \pi -\sin \left ( -\displaystyle \frac{\pi }{5} \right ) \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{\pi }{5}\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5}}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\cos \displaystyle \frac{\pi }{5}\sin \displaystyle \frac{\pi }{5} \right )}{4\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\left (\sin \displaystyle \frac{2\pi }{5}-\sin 0 \right )}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\cos \displaystyle \frac{3\pi }{5}\sin \displaystyle \frac{2\pi }{5}}{8\sin \displaystyle \frac{\pi }{5}}\\ &=\displaystyle \frac{\sin \pi -\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{\pi }{5}}{16\sin \displaystyle \frac{\pi }{5}}\\ &=\color{red}-\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 98.&\textrm{Nilai dari}\: \: \sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{16}&&\textrm{d}.\quad \displaystyle \frac{2}{16}\\\\ \textrm{b}.\quad \displaystyle \frac{4}{16}&\textrm{c}.\quad \displaystyle \frac{3}{16}&\textrm{e}.\quad \color{red}\displaystyle \frac{1}{16} \end{array}\\\\ &\textbf{(Olimpiade Sains PORSEMA NU 2012)}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\\ &=\displaystyle \frac{1}{4}\left ( 2\sin \displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24} \right )\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos \left ( \frac{10\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right )\times \left ( \cos \left ( \frac{2\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos 75^{\circ}-\cos 90^{\circ} \right )\times \left ( \cos 15^{\circ}-\cos 90^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \cos 75^{\circ}.\cos 15^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left [ \cos 90^{\circ}+\cos 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left ( 0+\frac{1}{2} \right )\\ &=\color{red}\displaystyle \frac{1}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 99.&\textrm{Nilai dari}\\ & \qquad\qquad\sin 18^{\circ}\cos 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{6} &&&\textrm{d}.&\displaystyle \frac{1}{3}\\\\ \textrm{b}.&\displaystyle \frac{1}{5}&\quad \textrm{c}.&\displaystyle \color{red}\displaystyle \frac{1}{4}\quad &\textrm{e}.&\displaystyle \frac{1}{2} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin 18^{\circ}\cos 36^{\circ}\\ &=\sin 18^{\circ}\cos 36^{\circ}\times \displaystyle \frac{2\cos 18^{\circ}}{2\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\left ( \sin 36^{\circ}+\sin 0^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 36^{\circ}\sin 36^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin 72^{\circ}}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\sin \left ( 90^{\circ}-18^{\circ} \right )}{4\cos 18^{\circ}}\\ &=\displaystyle \frac{\cos 18^{\circ}}{4\cos 18^{\circ}}\\ &=\color{red}\displaystyle \frac{1}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 100.&\textrm{Nilai eksak dari}\: \: \sin 36^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \displaystyle \frac{1}{4}\sqrt{10+2\sqrt{5}}&&&\textrm{d}.&\displaystyle \frac{\sqrt{5}-1}{4}\\ \textrm{b}.&\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}}&&\quad &\textrm{e}.&\displaystyle \frac{\sqrt{5}-1}{2}\\ \textrm{c}.&\displaystyle \displaystyle \frac{\sqrt{5}+1}{4} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$.

$.\qquad\begin{aligned}&\textrm{Perhatikan bahwa}\: \: \color{red}\bigtriangleup ABC\: \: \color{black}\textrm{sama kaki}\\ &\textrm{dengan}\: \: AD=DC=CB=1,\: AC=x\\ &\textrm{Diketahui pula}\: \: CD\: \: \textrm{adalah garis bagi}\\ &\textrm{serta}\: \: ABC\: \: \textrm{sebangun}\: \: \bigtriangleup BCD\\ &\textrm{akibatnya}:\\ &\color{red}\textrm{perbandingan sisi yang bersesuaian}\\ &\color{red}\textrm{akan sama},\: \: \color{black}\textrm{maka}\\ &\displaystyle \frac{AB}{BC}=\displaystyle \frac{BC}{AB-AD}\\ &\Leftrightarrow \displaystyle \frac{x}{1}=\frac{1}{x-1}\\ &\Leftrightarrow x(x-1)=1\\ &\Leftrightarrow x^{2}-x-1=0\\ &\Leftrightarrow x=\displaystyle \frac{1\pm \sqrt{5}}{2}\\ &\textrm{akibatnya}\: \: AB=AC=\displaystyle \frac{1+\sqrt{5}}{2}\\ &\textrm{Selanjutnya gunakan}\: \: \color{blue}\textrm{aturan sinus}\\ &\displaystyle \frac{AB}{\sin \angle C}=\frac{BC}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{AB}{BC}=\frac{\sin \angle C}{\sin \angle A}\\ &\Leftrightarrow \displaystyle \frac{ \left (\displaystyle \frac{1+\sqrt{5} }{2} \right )}{1}=\frac{\sin 72^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=\displaystyle \frac{2\sin 36^{\circ}\cos 36^{\circ}}{\sin 36^{\circ}}\\ &\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{2}=2\cos 36^{\circ}\\ &\Leftrightarrow \cos 36^{\circ}=\Leftrightarrow \displaystyle \frac{1+\sqrt{5}}{4}\\ &\textrm{Dari fakta di atas kita akan dengan}\\ &\textrm{mudah menentukan nilai sinusnya}\\ &\textrm{yaitu dengan menggunakan}\\ &\textrm{identitas trigonometri berikut}:\\ &\sin ^{2}36^{\circ}+\cos ^{2}36^{\circ}=1\\ &\Leftrightarrow \sin ^{2}36^{\circ}=1-\cos ^{2}36^{\circ}\\ &\Leftrightarrow \sin 36^{\circ}=\sqrt{1-\cos ^{2}36^{\circ}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\left ( \displaystyle \frac{1+\sqrt{5}}{4} \right )^{2}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{1-\displaystyle \frac{6+2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\sqrt{\displaystyle \frac{10-2\sqrt{5}}{16}}\\ &\Leftrightarrow \: \: \quad\quad\quad =\color{red}\displaystyle \frac{1}{4}\sqrt{10-2\sqrt{5}} \end{aligned}$.

 

Latihan Soal 9 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 81.&\textrm{Nilai}\: \: \cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-\displaystyle \frac{1}{2}\sqrt{6}&&&\textrm{d}.&\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&-\displaystyle \frac{1}{2}\sqrt{3}&\textrm{c}.&\color{red}-\displaystyle \frac{1}{2}\sqrt{2}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\cos \displaystyle \frac{5}{12}\pi -\cos \displaystyle \frac{1}{12}\pi \\ &=-2\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi +\displaystyle \frac{1}{12}\pi }{2} \right )\sin \left ( \displaystyle \frac{\displaystyle \frac{5}{12}\pi -\displaystyle \frac{1}{12}\pi }{2} \right )\\ &=-2\sin \left (\displaystyle \frac{\displaystyle \frac{6}{12}\pi }{2} \right )\sin \left (\displaystyle \frac{\displaystyle \frac{4}{12}\pi }{2} \right ) \\ &=-2\sin \left (\displaystyle \frac{1 }{4}\pi \right )\sin \left (\displaystyle \frac{1 }{6}\pi \right ) \\ &=-2\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{red}\displaystyle -\frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 82.&\textrm{Bentuk}\: \: \sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -2\sin 3x.\sin x &&&\textrm{d}.&\color{red}\displaystyle 2\sin 3x.\sin x\\ \textrm{b}.&\displaystyle -2\cos 3x.\sin x&&&\textrm{e}.&\displaystyle 2\cos 3x.\sin x\\ \textrm{c}.&\displaystyle 2\sin 2\left ( x-\pi \right ) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\sin \left ( 2x-\displaystyle \frac{3}{2}\pi \right )-\sin \left ( 4x+\displaystyle \frac{1}{2}\pi \right )\\ &=2\cos \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi+4x+\displaystyle \frac{1}{2}\pi}{2} \right )\\ &\qquad \times \sin \left ( \displaystyle \frac{2x-\displaystyle \frac{3}{2}\pi-\left (4x+\displaystyle \frac{1}{2}\pi \right )}{2} \right )\\ &=2\cos (3x-\displaystyle \frac{1}{2}\pi )\sin\left ( -x-\pi \right )\\ &=2\cos \left ( \displaystyle \frac{1}{2}\pi -3x \right )\left ( -\sin (\pi +x) \right )\\ &=2\left ( \sin 3x \right )\left ( -(-\sin x) \right )\\ &=2\sin 3x.\sin x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 83.&\textrm{Nilai}\: \: \displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}-1&&\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{6}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.\quad 1&\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{a}\\ &\begin{aligned}&\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{-2\sin \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}{2\cos \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}\\ &=-\displaystyle \frac{2\sin 45^{\circ}\times \sin 30^{\circ}}{2\cos 45^{\circ}\times \sin 30^{\circ}}\\ &=-\tan 45^{\circ}\\ &=\color{red}-1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 84.&\textrm{Bentuk}\: \: \displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ & \textrm{senilai dengan}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\tan 6x &&&\textrm{d}.&6\cot x\\ \textrm{b}.&\displaystyle -\cot 6x&&&\textrm{e}.&\displaystyle \color{red}\tan 6x\\ \textrm{c}.&\displaystyle 6\tan x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}\\ &=\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}\\ &=\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}\\ &=\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}\\ &=\tan 6x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 85.&\textrm{Nilai dari}\\ &\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan 2x &&\textrm{d}.&\displaystyle \tan 8x\\ \textrm{b}.&\color{red}\displaystyle \tan 4x&\quad&\textrm{e}.&\displaystyle \tan 16x\\ \textrm{c}.&\displaystyle \displaystyle \tan 6x \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}\\ &=\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}\\ &=\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}\\ &=\displaystyle \frac{2\sin 4x\left ( \cos 3x+\cos x \right )}{2\cos 4x\left ( \cos 3x+\cos x \right )}\\ &=\tan 4x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 86.&\textrm{Bentuk sederhana dari}\\ &\quad\quad \displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \tan A &&&\textrm{d}.&2\cos 2A\\ \textrm{b}.&\displaystyle 2\tan A&&&\textrm{e}.&\displaystyle \color{red}2\tan 2A\\ \textrm{c}.&\displaystyle 2\sin 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}\\ &=\displaystyle \frac{(\cos A+\sin A)^{2}-(\cos A-\sin A)^{2}}{(\cos A-\sin A)(\cos A+\sin A)}\\ &=\displaystyle \frac{(\cos ^{2}A+2\cos A\sin A+\sin ^{2}A)-(\cos ^{2}A-2\cos A\sin A+\sin ^{2}A)}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{4\cos A\sin A}{\cos ^{2}A-\sin ^{2}A}\\ &=\displaystyle \frac{2\sin 2A}{\cos 2A}\\ &=2\tan 2A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 87.&\textrm{Bentuk sederhana dari}\\ &\qquad\quad \displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\sin (x-y) &&&\textrm{d}.&\cos (x-y)\\ \textrm{b}.&\displaystyle \color{red}-\tan (x-y)&&&\textrm{e}.&\displaystyle \tan (x-y)\\ \textrm{c}.&\displaystyle \sin (x+y) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}\\ &=\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}\\ &=-\tan (x-y) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 88.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad \color{red}0 \end{array}\\\\ &\textbf{Jawab}:\: \: \color{red}\textbf{e}\\ &\begin{aligned}&\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}\\ &=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=\color{red}0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 89.&\textrm{Nilai dari}\\ &\quad\quad 8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle 4(\sqrt{3}+\sqrt{2}) &&&\textrm{d}.&\color{red}2(\sqrt{3}-\sqrt{2})\\ \textrm{b}.&\displaystyle 4(\sqrt{3}-\sqrt{2})&&&\textrm{e}.&\displaystyle \sqrt{3}-\sqrt{2}\\ \textrm{c}.&\displaystyle 2(\sqrt{3}+\sqrt{2}) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&8\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times 2\cos 82,5^{\circ}\sin 37,5^{\circ}\\ &=4\times \left ( \sin (82,5^{\circ}+37,5^{\circ})-\sin (82,5^{\circ}-37,5^{\circ}) \right )\\ &=4\times \left (\sin 120^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin \left ( 180^{\circ}-60^{\circ} \right )-\sin 45^{\circ} \right )\\ &=4\times \left ( \sin 60^{\circ}-\sin 45^{\circ} \right )\\ &=4\times \left ( \displaystyle \frac{1}{2}\sqrt{3}-\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=2\left ( \sqrt{3}-\sqrt{2} \right ) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 90.&\textrm{Bentuk lain dari}\\ &\quad\quad -2\cos 5A.\cos 7A\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\cos 6A-\cos A &&&\\ \textrm{b}.&\displaystyle -\cos 6A+\cos A&&&\\ \textrm{c}.&\displaystyle \cos 12A-\cos 2A\\ \textrm{d}.&-\cos 12A+\cos 2A\\ \textrm{e}.&\color{red}\displaystyle -\cos 12A-\cos 2A \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}&-2\cos 5A.\cos 7A\\ &=-\left ( 2\cos 5A.\cos 7A \right )\\ &=-\left ( \cos 12A+\cos (-2A) \right )\\ &=-\left ( \cos 12A+\cos 2A \right )\\ &=-\cos 12A-\cos 2A \\ \end{aligned} \end{array}$.

Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 71.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\cos x \\\\ &\textrm{b}.\quad \displaystyle y=\cos 2x \\\\ &\textrm{c}.\quad \displaystyle y=\cos \displaystyle \frac{1}{2}x \\\\ &\textrm{d}.\quad \displaystyle \color{red}y=2\cos 2x \\\\ &\textrm{e}.\quad \displaystyle y=2\cos \displaystyle \frac{1}{2}x \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\\ & \textbf{atas dan ke bawah}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{2}=180^{\circ}=\pi ,\\ & \textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\cos 2x \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 72.&\textrm{Grafik fungsi trigonometri pada gambar}\\ &\textrm{berikut adalah}\: .... \end{array}$.

$.\quad\begin{array}{ll}\\ &\textrm{a}.\quad \displaystyle y=2\sin \left (2x+\pi \right ) \\\\ &\textrm{b}.\quad \displaystyle y=\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{c}.\quad \displaystyle y=2\sin \left (2x-\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{d}.\quad \displaystyle y=\sin \left (2x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textrm{e}.\quad \displaystyle \color{red}y=2\sin \displaystyle \left (x+\displaystyle \frac{1}{2}\pi \right ) \\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Dari grafik tampak jelas bahwa}\\ &\textrm{gambar di atas adalah garfik}\\ &\textrm{fungsi cosinus di geser ke}\: \: \textbf{kiri}\\ &\textrm{dengan}\: \: \textbf{amplitudo}\: \: 2\\ &\textrm{dan}\: \: \textbf{periodenya}\: \: \displaystyle \frac{360^{\circ}}{1}=360^{\circ}=2\pi ,\\ &\textrm{maka}\\ &\textrm{bentuk persamaan}\: \: \textbf{grafik fungsinya}\\ &y=\color{red}2\sin \left ( x+\displaystyle kx \right )\\ &\textrm{dengan}\: \: +k\: \: \textrm{adalah}\\ &\textbf{besar geseran ke kiri}\: \: \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: 90^{\circ}\\ &\textrm{Jadi},\: \: y=\color{red}2\sin (x+\displaystyle \frac{1}{2}\pi )^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 73.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\sin x=\sin \displaystyle \frac{2}{10}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{2}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{22}{10}\pi ,\displaystyle \frac{28}{10}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{12}{10}\pi ,\displaystyle \frac{8}{10}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \sin x=\sin \displaystyle \frac{2}{10}\pi \\ &\Leftrightarrow \: \: x_{1}=\displaystyle \frac{2}{10}\pi+k.2\pi \: \: \: \: \color{blue}\textrm{atau}\\ &\Leftrightarrow \quad x_{2} =\left (\pi -\displaystyle \frac{2}{10}\pi \right )+k.2\pi=\displaystyle \frac{8}{10}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{2}{10}\pi\: \: (\color{blue}\textrm{mm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{8}{10}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1,2}=....+2\pi \quad (\color{red}\textrm{tidak memenuhi})\\ \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{2}{10}\pi,\: \displaystyle \frac{8}{10}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 74.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \: \: \textrm{untuk}\: \: 0^{\circ}\leq x\leq 360^{\circ}\\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \displaystyle \left \{ \displaystyle \frac{1}{3}\pi ,\pi ,\displaystyle \frac{5}{3}\pi ,\displaystyle \frac{7}{3}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{1}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{2}{4}\pi ,\displaystyle \frac{3}{4}\pi ,\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \color{red}\left \{ \displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,\displaystyle \frac{7}{4}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \tan \left ( 2x-\displaystyle \frac{1}{4}\pi \right )=\tan \displaystyle \frac{1}{4}\pi \\ &\Leftrightarrow \: \: 2x-\displaystyle \frac{1}{4}\pi=\displaystyle \frac{1}{4}\pi+k.\pi\\ &\Leftrightarrow \quad 2x =\displaystyle \frac{2}{4}\pi +k.\pi \\ &\Leftrightarrow \quad x =\displaystyle \frac{1}{4}\pi +k.\frac{\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{1}{4}\pi\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{\pi}{2}=\displaystyle \frac{3}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=2\Rightarrow x=\displaystyle \frac{1}{4}\pi+\pi =\displaystyle \frac{5}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=3\Rightarrow x=\displaystyle \frac{1}{4}\pi+\frac{3\pi}{2}=\displaystyle \frac{7}{4}\pi \: \: (\color{blue}\textrm{mm}) \\ &k=4\Rightarrow x=\displaystyle \frac{1}{4}\pi+2\pi =\displaystyle \frac{9}{4}\pi \: \: (\color{red}\textrm{tidak memenuhi}) \\ \end{aligned} \\ &\textbf{HP}=\color{red}\color{red}\left \{\displaystyle \frac{1}{4}\pi,\: \displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 75.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\cos 2x-2\cos x=-1\: \: \textrm{untuk}\: \: 0< x< 2\pi \\ &\textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{3}{2}\pi, 2\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi, 2\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi ,\displaystyle \frac{3}{2}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\displaystyle \frac{2}{3}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ 0,\displaystyle \frac{1}{2}\pi ,\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & \cos 2x-2\cos x=-1\\ &\Leftrightarrow \cos 2x-2\cos x+1=0\\ &\Leftrightarrow \left ( 2\cos ^{2}x-1 \right )-2\cos x+1=0 \\ &\Leftrightarrow 2\cos x\left ( \cos x-1 \right )=0\\ &\Leftrightarrow \cos x=0\: \: \textrm{atau}\: \: \cos x=1\\ &\Leftrightarrow \cos x=\cos \displaystyle \frac{1}{2}\pi \: \: \textrm{atau}\: \: \cos x=\cos 0\\ &\Leftrightarrow x_{1,2}=\pm \displaystyle \frac{1}{2}\pi +k.2\pi \: \: \textrm{atau}\: \: x_{3}=k.2\pi\\ &\textrm{maka}\\ &k=0\Rightarrow x_{1}=-\displaystyle \frac{1}{2}\pi\: \: (\color{red}\textrm{tm})\: \: \color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=\displaystyle \frac{1}{2}\pi\: \: (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{3}=\displaystyle 0\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x_{1}=\displaystyle \frac{3}{2}\pi \quad (\color{blue}\textrm{mm})\\ &\qquad\qquad x_{2}=\displaystyle \frac{5}{2}\pi\: \: (\color{red}\textrm{tm})\\ &\qquad\qquad x_{3}=\displaystyle 2\pi\: \: (\color{blue}\textrm{mm})\\ \end{aligned} \end{array} \end{array}$.

$\begin{array}{ll}\\ 76.&\textrm{Himpunan penyelesaian dari persamaan}\\ &\qquad\quad 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3} \\ & \textrm{untuk}\: \: 0\leq x\leq \pi \: \: \: \textrm{adalah}\: ....\\\\ &\textrm{a}.\quad \color{red}\displaystyle \left \{ \displaystyle \frac{1}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{b}.\quad \displaystyle \left \{ \displaystyle \frac{2}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{c}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\textrm{d}.\quad \displaystyle \left \{ \displaystyle \frac{3}{12}\pi ,\displaystyle \frac{9}{12}\pi \right \} \\\\ &\textrm{e}.\quad \displaystyle \left \{ \displaystyle \frac{5}{12}\pi ,\displaystyle \frac{7}{12}\pi \right \} \\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\begin{aligned} & 3\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\sqrt{3}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\frac{\sqrt{3}}{3}\\ &(\textrm{kuadran IV, karena Y negatif, X positif})\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=-\tan \displaystyle \frac{1}{6}\pi ,\: \: \textbf{menjadi}\\ &\tan \left (2x-\displaystyle \frac{1}{3}\pi \right )=\tan \left ( 2\pi -\displaystyle \frac{1}{6}\pi \right )=\tan \displaystyle \frac{11}{6}\pi \\ &\left (2x-\displaystyle \frac{1}{3}\pi \right )=\displaystyle \frac{11}{6}\pi\\ &\Leftrightarrow \: \: 2x=\displaystyle \frac{1}{3}\pi+\displaystyle \frac{11}{6}\pi +k.\pi =\displaystyle \frac{13}{6}\pi +k.\pi \\ &\Leftrightarrow \: \: x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{k.\pi}{2} \\ &k=0\Rightarrow x=\displaystyle \frac{13}{12}\pi=\displaystyle \frac{1}{12}\pi \: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=\displaystyle \frac{13}{12}\pi +\displaystyle \frac{1}{2}\pi =\displaystyle \frac{19}{12}\pi=\displaystyle \frac{7}{12}\pi \quad (\color{blue}\textrm{mm})\\ &k=2\Rightarrow x=\displaystyle \frac{13}{12}\pi +\pi \quad \color{red}\textrm{tidak memenuhi} \end{aligned} \\ &\textbf{HP}=\color{red}\left \{\displaystyle \frac{1}{12}\pi,\: \displaystyle \frac{7}{12}\pi \right \} \end{array} \end{array}$.

$\begin{array}{ll}\\ 77.&\textrm{Salah satu nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \cos x+\sin x=\displaystyle \frac{1}{2}\sqrt{6}\\ & \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{24}\pi &&&\textrm{d}.&\displaystyle \frac{1}{8}\pi \\\\ \textrm{b}.&\displaystyle \frac{1}{15}\pi&\textrm{c}.&\displaystyle \color{red}\frac{1}{12}\pi&\textrm{e}.&\displaystyle \frac{1}{6}\pi \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui bahwa}\\ &\sin x+\cos x=\displaystyle \frac{1}{2}\sqrt{6}\quad \left (\textbf{ingat}:a=1,\: b=1 \right )\\ &\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\begin{cases} k & =\sqrt{1^{2}+1^{2}}=\sqrt{2} \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{1}{1}=1\Rightarrow \theta =45^{\circ}=\displaystyle \frac{1}{4}\pi \end{cases}\\ &\qquad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran I, karena}\: \: a,b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&\sin x+\cos x=k\cos \left ( x-\theta \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \sqrt{2}\cos\left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{1}{2}\sqrt{6}\\ &\Leftrightarrow \: \: \, \cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{6}}{\sqrt{2}}=\displaystyle \frac{1}{2}\sqrt{3}\\ &\Leftrightarrow \: \: \,\cos \left ( x-\displaystyle \frac{1}{4}\pi \right )=\cos \displaystyle \frac{1}{6}\pi \\ &\Leftrightarrow \quad x-\displaystyle \frac{1}{4}\pi =\pm \displaystyle \frac{1}{6}\pi +k.2\pi \\ &\Leftrightarrow \quad x=\displaystyle \frac{1}{4}\pi\pm \displaystyle \frac{1}{6}\pi+k.2\pi \\ &\Leftrightarrow \quad x_{1}=\displaystyle \frac{5}{12}\pi+k.2\pi \: \: \textbf{atau}\\ &\: \: \: \quad\quad x_{2}=\displaystyle \frac{1}{12}\pi +k.2\pi \\ &k=0\Rightarrow x_{1}=\displaystyle \frac{5}{12}\pi\qquad (\color{blue}\textrm{memenuhi})\\ &\: \: \qquad\Rightarrow x_{2}=\displaystyle \color{red}\frac{1}{12}\pi\qquad \color{black}(\color{blue}\textrm{memenuhi})\\ &\textrm{Langkah berikutnya tidak diperlukan}\\ &\textrm{karena jawaban sudah kita dapatkan}\\ &\textrm{yaitu}:\: \: \color{red}\displaystyle \frac{1}{12}\pi \end{aligned} \end{array} \end{array}$.

$\begin{array}{ll}\\ 78.&\textrm{Himpunan penyelesaian persamaan}\\ &\qquad\: \: \cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1\\ &\textrm{untuk}\: \: 0\leq x< 360\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \color{red}\left \{ 0,240 \right \} &&&\textrm{d}.&\displaystyle \left \{ 180,240 \right \} \\\\ \textrm{b}.&\displaystyle \left \{ 150,270 \right \}&\textrm{c}.&\displaystyle \left \{ 180,300 \right \}&\textrm{e}.&\displaystyle \left \{ 210,270 \right \} \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{array}{ll}\\ &\textrm{Diketahui dari soal bahwa}\\ &\cos x^{\circ}-\sqrt{3}\sin x^{\circ}=1,\\ &\textrm{lalu kita ubah posisinya menjadi}\\\\ &-\sqrt{3}\sin x+\cos x=1\: \: \left (\textbf{ingat}:a=-\sqrt{3},\: b=1 \right )\\ &-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\begin{cases} k & =\sqrt{\left ( -\sqrt{3} \right )^{2}+\left ( 1 \right )^{2}}=\sqrt{4}=2 \\ \tan & \theta =\displaystyle \frac{a}{b}=\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta =300^{\circ} \end{cases}\\ &\quad\quad\textrm{sudut}\: \: \theta \: \: \textrm{di kuadran IV, karena}\: \: a<0,\: b>0\\ &\textrm{selanjutnya}\\ &\begin{aligned}&-\sqrt{3}\sin x+\cos x=k\cos \left ( x-\theta \right )=1\\ &\Leftrightarrow 2\cos\left ( x-300^{\circ} \right )=1\\ &\Leftrightarrow \: \: \, \cos \left ( x-300^{\circ} \right )=\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \: \,\cos \left ( x-300^{\circ} \right )=\cos 60^{\circ}\\ &\Leftrightarrow \quad x-300^{\circ} =\pm 60^{\circ}+k.360^{\circ}\\ &\Leftrightarrow \quad x=300^{\circ}\pm 60^{\circ}+k.360^{\circ}\\ &k=0\Rightarrow x_{1}=300^{\circ}+60^{\circ}=360^{\circ}=0^{\circ}\: \: (\color{blue}\textrm{mm})\\ & \qquad\qquad\quad\color{black}\textrm{atau}\\ &\qquad\qquad x_{2}=300^{\circ}-60^{\circ}=240^{\circ}\: \: (\color{blue}\textrm{mm})\\ &k=1\Rightarrow x=300^{\circ}\pm 60^{\circ}+360^{\circ}\quad (\color{red}\textrm{tm})\\ \end{aligned} \\ &\textbf{HP}=\left \{0^{\circ},240^{\circ} \right \} \end{array} \end{array}$

$\begin{array}{ll}\\ 79.&\textrm{Diketahui}\: \: \alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah sudut} \: \: \textbf{lancip}\\ &\textrm{Nilai dari}\: \: \cos \left ( \alpha +\beta \right )\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&1&&&\textrm{d}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{1}{2}&\textrm{e}.&\color{red}\displaystyle 0 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bullet \quad\alpha -\beta =\displaystyle \frac{\pi }{3}\: \: \textrm{dan}\: \: \sin \alpha \sin \beta =\displaystyle \frac{1}{4}\\ &\bullet \quad\textrm{dengan}\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{sudut} \: \: \textbf{lancip}\\ &\qquad \textrm{akibatnya semua sudut dikuadran I}\\ &\qquad \textrm{sehingga}\color{purple}\begin{cases} \sin & =+ \\ \cos & =+ \\ \tan & =+ \end{cases}\\ &\textrm{ditanya}\: \: \cos \left ( \alpha +\beta \right ),\: \: \textrm{maka}\\ &\textrm{sebagai langkah awal kita adalah}:\\ &\cos \left ( \alpha -\beta \right )=\cos \left ( \displaystyle \frac{\pi }{3} \right )\\ &\Leftrightarrow \: \cos \alpha \cos \beta +\sin \alpha \sin \beta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta+\displaystyle \frac{1}{4} =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \cos \alpha \cos \beta =\displaystyle \frac{1}{2}-\displaystyle \frac{1}{4}=\displaystyle \frac{1}{4}\\ &\textbf{Selanjutnya nilai dari}\\ &\cos \left ( \alpha +\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &=\displaystyle \frac{1}{4}-\displaystyle \frac{1}{4}=0\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 80.&\textrm{Nilai}\: \: \sin 75^{\circ}-\sin 165^{\circ}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\sqrt{2}&&&\textrm{d}.&\color{red}\displaystyle \frac{1}{2}\sqrt{2}\\\\ \textrm{b}.&\displaystyle \frac{1}{4}\sqrt{3}&\textrm{c}.&\displaystyle \frac{1}{4}\sqrt{6}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{6} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\sin 75^{\circ}-\sin 165^{\circ}\\ &=2\cos \left ( \displaystyle \frac{75^{\circ}+165^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-165^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{240^{\circ}}{2}\sin \left (-\displaystyle \frac{90^{\circ}}{2} \right ) \\ &=2\cos 120^{\circ}\sin \left ( -45^{\circ} \right )\\ &=2\left (-\cos 60^{\circ} \right )\left (- \sin 45^{\circ} \right )\\ &=2\left ( -\displaystyle \frac{1}{2} \right )\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.

Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 61.&\textrm{Nilai}\: \: 105^{\circ}\: \: \textrm{jika dinyatakan ke radian}\\ &\textrm{adalah}\: \: ....\: \: \textrm{radian}\\\\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\pi \\\\ &\textrm{b}.\quad \displaystyle \frac{5}{6}\pi \\\\ &\textrm{c}.\quad \displaystyle \frac{5}{12}\pi \\\\ &\textrm{d}.\quad \color{red}\displaystyle \frac{7}{12}\pi \\\\ &\textrm{e}.\quad \displaystyle \frac{9}{12}\pi \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Diketah}&\textrm{ui bahwa}\\ 180^{\circ}&=\pi \: \: \: radian\\ 1^{\circ}&=\displaystyle \frac{\pi }{180}\: \: \: radian\\ 105\times 1^{\circ}&=105\times \displaystyle \frac{\pi }{180}\: \: \: radian\\ 105^{\circ}&=\displaystyle \frac{7}{12}\pi \: \: \: radian \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 62.&\textrm{Nilai}\: \: \tan 240^{\circ} \: \: \: \textrm{adalah}\: \: ....\\\\ &\textrm{a}.\quad \displaystyle \color{red}\sqrt{3} \\\\ &\textrm{b}.\quad \displaystyle \frac{1}{3}\sqrt{3} \\\\ &\textrm{c}.\quad \displaystyle -\frac{1}{3}\sqrt{3} \\\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\sqrt{3} \\\\ &\textrm{e}.\quad \displaystyle -\sqrt{3} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\tan 240^{\circ}&=\tan \left ( 180^{\circ}+60^{\circ} \right )\\ &=\tan 60^{\circ}\\ &=\color{red}\sqrt{3}\\ \textbf{catatan}&: \textrm{ingat sudut berelasi} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 63.&\textrm{Perhatikanlah gambar berikut}\\ \end{array}$.

$.\qquad\begin{array}{ll}\\ &\textrm{Pada gambar di atas perbandingan}\\ &\sin \theta \: \: \textrm{adalah}\: ....\\ &\textrm{a}.\quad \sqrt{\displaystyle \frac{a^{2}-d^{2}}{f^{2}+g^{2}}} \\ &\textrm{b}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}+g^{2}}} \\ &\textrm{c}.\quad \sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}-g^{2}}} \\ &\textrm{d}.\quad \sqrt{\displaystyle \frac{a^{2}+b^{2}}{f^{2}-g^{2}}}\\ &\textrm{e}.\quad \color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Dari so}&\textrm{al diketahui bahwa}\\ \sin \theta &=\displaystyle \frac{c}{e}=\displaystyle \frac{\sqrt{a^{2}-b^{2}}}{\sqrt{f^{2}+g^{2}}}\\ &=\color{red}\sqrt{\displaystyle \frac{a^{2}-b^{2}}{f^{2}+g^{2}}} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 64.&\textrm{Nilai dari}\: \: \left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right ) \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}0&&&\textrm{d}.&1\\ \textrm{b}.&\displaystyle \frac{1}{3}&\textrm{c}.&\displaystyle \frac{2}{\sqrt{3}}&\textrm{e}.&\displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( \cos ^{2}17^{\circ}-\sin ^{2}73^{\circ} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin 73^{\circ} \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\left (\sin \left ( 90^{\circ}-17^{\circ} \right ) \right )^{2} \right )\\ &=\left ( \cos ^{2}17^{\circ}-\cos ^{2}17^{\circ} \right )\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 65.&\textrm{Jika diketahui}\\ & \displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ},\\ & \textrm{maka nilai}\: \: x\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-2&&&\textrm{d}.&\color{red}1\\ \textrm{b}.&\displaystyle -1&\textrm{c}.&0&\textrm{e}.&\displaystyle 2 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{x\csc ^{2}30^{\circ}\sec ^{2}45^{\circ}}{8\cos ^{2}45^{\circ}\sin ^{2}60^{\circ}}=\tan ^{2}60^{\circ}-\tan ^{2}30^{\circ}\\ &\displaystyle \frac{x\left ( 4 \right )\left ( \displaystyle \frac{4}{2} \right )}{8\left ( \displaystyle \frac{2}{4} \right )\left ( \displaystyle \frac{3}{4} \right )}=3-\left ( \displaystyle \frac{1}{3} \right )\\ &\displaystyle \frac{8x}{3}=\displaystyle \frac{8}{3}\\ &x=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 66.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&0,143\\ \textrm{b}.&\displaystyle -0,321&\textrm{c}.&\color{red}0&\textrm{e}.&\displaystyle 0,321 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\displaystyle \frac{\sin 49^{\circ}}{\cos 41^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\sin 73^{\circ}}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\cos \left (90^{\circ}-49^{\circ} \right )}-\displaystyle \frac{\cos 17^{\circ}}{\sin \left (90^{\circ}-17^{\circ} \right )}\\ &=\displaystyle \frac{\sin 49^{\circ}}{\sin 49^{\circ}}-\displaystyle \frac{\cos 17^{\circ}}{\cos 17^{\circ}}\\ &=1-1\\ &=0 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 67.&\textrm{Nilai dari}\\ &p=r\sin \alpha \cos \beta \\ &q=r\sin \alpha \sin \beta \\ &s=r\cos \alpha \\ &\textrm{maka pernyataan berikut yang}\\ &\textrm{tepat adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\color{red}p^{2}+t^{2}+s^{2}=r^{2}&&&\\ \textrm{b}.&p^{2}-t^{2}+s^{2}=r^{2} \\ \textrm{c}.&p^{2}+t^{2}-s^{2}=r^{2}&\\ \textrm{d}.&-p^{2}+t^{2}+s^{2}=r^{2}&\\ \textrm{e}.&-p^{2}-t^{2}+s^{2}=r^{2} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Saat}\\ &p^{2}+q^{2}\: \: \textrm{maka hasilnya adalah}\\ &\color{purple}\begin{array}{lll}\\ p^{2}&=r^{2}\sin^{2} \alpha \cos^{2} \beta&\\ q^{2}&=r^{2}\sin^{2} \alpha \sin^{2} \beta&+\\\hline &=r^{2}\sin ^{2}\alpha \left ( \cos ^{2}\beta +\sin ^{2}\beta \right )\\ &=r^{2}\sin ^{2}\alpha (1)\\ &=r^{2}\sin ^{2}\alpha \end{array}\\ &\textrm{Dan saat}\\ &p^{2}+q^{2}+s^{2}\: \: \textrm{akan diperoleh hasil}\\ &\color{purple}\begin{array}{lll}\\ p^{2}+q^{2}&=r^{2}\sin ^{2}\alpha &\\ \qquad s^{2}&=r^{2}\cos ^{2}\alpha &+\\\hline &=r^{2}\sin ^{2}\alpha+r^{2}\cos ^{2}\alpha\\ &=r^{2}\left ( \sin ^{2}\alpha+\cos ^{2}\alpha \right )\\ &=r^{2}(1)\\ &=r^{2} \end{array}\\ \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 68.&\textrm{Nilai dari}\\ & \displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &\textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&-1&&&\textrm{d}.&-\tan \theta \\ \textrm{b}.&\displaystyle 0&\textrm{c}.&\color{red}1&\textrm{e}.&\displaystyle \tan \theta \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\color{purple}\textrm{Ingat kembali sudut-sudut}\\ &\color{purple}\textrm{yang berelasi dari kudran selain I}\\ &\color{purple}\textrm{ke kuadran I beserta tandanya}\\ &\displaystyle \frac{\cos \left ( 90^{\circ}+\theta \right )\sec \left ( 2\pi -\theta \right )\tan \left ( \pi -\theta \right )}{\sec \left ( \theta -2\pi \right )\sin \left ( 540^{\circ}+\theta \right )\cot \left ( \theta -90^{\circ} \right )}\\ &=\displaystyle \frac{\left (-\sin \theta \right ) .\sec \theta .\left (-\tan \theta \right )}{\sec \theta .\left (-\sin \theta \right ). \left (-\tan \theta \right )}\\ &=1 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 69.&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2} \\ &\textrm{maka nilai dari}\\ &\sin ^{3}\theta +\cos ^{3}\theta \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle \frac{1}{2}&&&\textrm{d}.&\displaystyle \frac{5}{8} \\\\ \textrm{b}.& \displaystyle \frac{3}{4}&\textrm{c}.&\displaystyle \frac{9}{15}&\textrm{e}.&\color{red}\displaystyle \frac{11}{16} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\sin \theta +\cos \theta =\displaystyle \frac{1}{2}\\ &\Leftrightarrow \: \left (\sin \theta +\cos \theta \right )^{2} =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 1+2\sin \theta \cos \theta =\displaystyle \frac{1}{4}\\ &\Leftrightarrow \: 2\sin \theta \cos \theta =-\displaystyle \frac{3}{4}\\ &\Leftrightarrow \: \sin \theta \cos \theta =-\displaystyle \frac{3}{8}\\ &\textbf{Selanjutnya}\\ &\color{purple}\sin ^{3}\theta +\cos ^{3}\theta\\ &=\color{purple}\left ( \sin \theta +\cos \theta \right )\left ( \sin ^{2}\theta -\sin \theta \cos \theta +\cos ^{2}\theta \right )\\ &=\color{purple}\left ( \displaystyle \frac{1}{2} \right )\left ( 1-\left ( -\displaystyle \frac{3}{8} \right ) \right ) \\ &=\color{purple}\displaystyle \frac{1}{2}\times \displaystyle \frac{11}{8}\\ &=\color{red}\displaystyle \frac{11}{16} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 70.&\textrm{Jika diketahui}\: \: \: \displaystyle \frac{3}{2}\pi <x<2\pi \\ &\textrm{dan}\: \: \: \tan x=m,\\ &\textrm{maka nilai dari}\: \: \sin x \cos x \: \: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll}\\ \textrm{a}.&\displaystyle -\frac{1}{m^{2}+1}&&&\textrm{d}.&\displaystyle -\frac{m}{m^{2}-1} \\\\ \textrm{b}.& \color{red}\displaystyle -\frac{m}{m^{2}+1}&\textrm{c}.&\displaystyle \frac{m}{m^{2}+1}&\textrm{e}.&\displaystyle \frac{m}{m^{2}-1} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\: \: \displaystyle \frac{3}{2}\pi <x<2\pi\\ &\textrm{ini daerah Kwadran IV, akibatnya adalah nilai}\\ &\begin{cases} \sin x & = -\\ \cos x & =+ \\ \tan x & =- \end{cases}\\ &\textbf{Selanjutnya ada pernyataan}\: \: \tan x=m\\ &\textrm{ini artinya}\: \: \tan x=\displaystyle \frac{m}{1}\\ &\textbf{Perhatikanlah ilustrasi gambar berikut} \end{aligned} \end{array}$.

$.\qquad\begin{aligned} &\textrm{maka nilai dari}\\ &\sin x\cos x\: \: \left (\textrm{ingat yang diminta di Kwadran IV} \right )\\ &=\left (-\displaystyle \frac{m}{\sqrt{m^{2}+1}} \right )\times \left (+\displaystyle \frac{1}{\sqrt{m^{2}+1}} \right )\\ &=\color{red}-\displaystyle \frac{m}{m^{2}+1} \end{aligned}$

Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

 $\begin{array}{ll}\\ 51.&\textrm{Nilai dari}\: \: \displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}=....\\ &\begin{array}{llll}\\ \textrm{a}.&-\infty \\ \textrm{b}.&\displaystyle -1\\ \textrm{c}.&\displaystyle 0\\ \color{red}\textrm{d}.&\displaystyle 1\\ {e}.&\displaystyle \infty \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{1}{\sec ^{2}A}+\frac{1}{\csc ^{2}A}\\ &=\cos ^{2}A+\sin ^{2}=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 52.&\textrm{Nilai dari}\: \: \displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....\\ &\begin{array}{llll}\\ \textrm{a}.&\cot B\times \cot C \\ \color{red}\textrm{b}.&\displaystyle \tan B\times \tan C\\ \textrm{c}.&\displaystyle \sec B\times \csc C\\ \textrm{d}.&\displaystyle \tan B\times \cot C\\ {e}.&\displaystyle \tan B\times \csc C \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}&\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}\\ &=\displaystyle \frac{\tan B+\tan C}{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}\\ &=\displaystyle \frac{\tan B+\tan C}{\left ( \displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C} \right )}\\ &=\color{red}\tan B\times \tan C \end{aligned} \end{array}$

$\begin{array}{ll}\\ 53.&\textrm{Nilai dari}\\ &\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....\\ &\begin{array}{llll}\\ \textrm{a}.&2\tan A \\ \textrm{b}.&2\cot A\\ \textrm{c}.&\displaystyle 2\sec A\\ \color{red}\textrm{d}.&\displaystyle 2\csc A\\ {e}.&\displaystyle 2\tan A.\sec A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}\\ &=\tan A\left (\displaystyle \frac{1}{\displaystyle \frac{1}{\cos A}-1}+\frac{1}{\displaystyle \frac{1}{\cos A}+1} \right )\\ &=\displaystyle \frac{\sin A}{\cos A}\left ( \displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A} \right )\\ &=\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}\\ &=\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}\\ &=\displaystyle \frac{2\sin A}{1-\cos ^{2}}\\ &=\displaystyle \frac{2\sin A}{\sin ^{2}A}\\ &=\displaystyle \frac{2}{\sin A}\\ &=\color{red}2\csc A \end{aligned}\\\\ &\textrm{Sebagai catatanya}\\ &\textrm{Anda bisa gunakan cara yang lain} \end{array}$

$\begin{array}{ll}\\ 54.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\sin \left ( 2x-20^{\circ} \right )=-\cos \left ( 3x+50^{\circ} \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-30^{\circ}\\ \textrm{b}.&-25^{\circ}\\ \color{red}\textrm{c}.&20^{\circ}\\ \textrm{d}.&25^{\circ}\\ {e}.&30^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\sin \left ( 2x-20^{\circ} \right )&=-\cos \left ( 3x+50^{\circ} \right )\\ \sin \left ( 20^{\circ}-2x \right )&=\cos \left ( 3x+50^{\circ} \right )\\ \sin A&=\cos B,\: \: \color{black}\textrm{artinya}\\ A+B&=90^{\circ},\: \: \color{magenta}\textrm{maka}\\ \left ( 20^{\circ}-2x \right )+\left ( 3x+50^{\circ} \right )&=90^{\circ}\\ x+70^{\circ}&=90^{\circ}\\ x&=90^{\circ}-70^{\circ}\\ &=\color{red}20^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 55.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\tan \left ( 2x+60^{\circ} \right )=\cot \left ( 90^{\circ}-3x \right )\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&20^{\circ}\\ \textrm{b}.&30^{\circ}\\ \textrm{c}.&40^{\circ}\\ \textrm{d}.&50^{\circ}\\ \color{red}\textrm{e}.&60^{\circ} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\tan \left ( 2x+60^{\circ} \right )&=\cot \left ( 90^{\circ}-3x \right )\\ \tan (2x+60^{\circ})&=\tan 3x\\ 2x+60^{\circ}&=3x\\ 2x-3x&=-60^{\circ}\\ -x&=-60^{\circ}\\ x&=\color{red}60^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 56.&\textrm{Jika nilai}\: \: \cot A+\cos A=x\\ &\textrm{dan}\: \: \cot A-\cos A=y,\\ &\textrm{maka nilai}\: \: \left ( x^{2}-y^{2} \right )=\: ....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&4\sqrt{xy}\\ \textrm{b}.&2\sqrt{xy}\\ \textrm{c}.&xy\\ \textrm{d}.&2\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}xy&=\left (\cot A+\cos A \right )\left ( \cot A-\cos A \right )\\ &=\cot ^{2}A-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\\ &=\displaystyle \frac{\cos^{2}A }{\sin ^{2}A}-\cos ^{2}A\times \frac{\sin ^{2}A}{\sin ^{2}A}\\ &=\displaystyle \frac{\cos ^{2}A}{\sin ^{2}A}\left ( 1-\sin ^{2}A \right )\\ &=\displaystyle \frac{\cos ^{4}A}{\sin ^{2}A}\\ \sqrt{xy}&=\displaystyle \frac{\cos A}{\sin A}\times \cos A\\ \color{black}\textrm{Se}&\color{black}\textrm{lanjutnya}\\ x^{2}-y^{2}&=\left (\cot A+\cos A \right )^{2}-\left ( \cot A-\cos A \right )^{2}\\ (x+y)&(x-y)=(\cot A+\cos A+\cot A-\cos A)\\ &\qquad\times (\cot A+\cos A-(\cot A-\cos A))\\ x^{2}-y^{2}&=2\cot A\times 2\cos A\\ &=4\times \displaystyle \frac{\cos A}{\sin A}\times \cos A\\ &=\color{red}4\sqrt{xy} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 57.&\textrm{Jika nilai}\: \: \cos A+\sin A=\sqrt{2}\cos A\\ &\textrm{maka nilai}\: \: \left ( \cos A-\sin A \right )=\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-\sqrt{2}\cos A\\ \textrm{b}.&-\sqrt{2}\sin A\\ \color{red}\textrm{c}.&\sqrt{2}\sin A\\ \textrm{d}.&\displaystyle \frac{1}{\sqrt{2}}\sec A\\ \textrm{e}.&\displaystyle \frac{1}{\sqrt{2}}\csc A \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\cos A+\sin A&=\sqrt{2}\cos A\\ \left (\cos A+\sin A \right )^{2}&=\left (\sqrt{2}\cos A \right )^{2}\\ 1+2\sin A\cos A&=2\cos ^{2}A\\ 2\sin A\cos A&=2\cos ^{2}A-1\\ \textrm{maka}&\\ \left (\cos A-\sin A \right )^{2}&=1-2\sin A\cos A\\ &=1-\left ( 2\cos ^{2}A-1 \right )\\ &=2-2\cos ^{2}A\\ &=2\left ( 1-\cos ^{2}A \right )\\ &=2\sin ^{2}A\\ \cos A-\sin A&=\sqrt{2\sin ^{2}A}\\ &=\sin A\sqrt{2}\\ &=\color{red}\sqrt{2}.\sin A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 58.&\textrm{Nilai}\: \: \cos \gamma \left ( \csc \gamma +\tan \gamma \right ) \\ & \textrm{adalah ekivalen dengan}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}\cot \gamma +\sin \gamma \\ \textrm{b}.&\tan \gamma +\cos \gamma \\ \textrm{c}.&\cot \gamma -\sin \gamma \\ \textrm{d}.&\tan \gamma -\cos \gamma \\ \textrm{e}.&\cot \gamma -\tan \gamma \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\cos \gamma \left ( \csc \gamma +\tan \gamma \right )&=\cos \gamma \left ( \displaystyle \frac{1}{\sin \gamma } +\displaystyle \frac{\sin \gamma }{\cos \gamma } \right )\\ &=\cos \gamma \left ( \displaystyle \frac{\cos \gamma +\sin ^{2}\gamma }{\sin \gamma \cos \gamma } \right )\\ &=\displaystyle \frac{\cos \gamma +\sin ^{2}\gamma }{\sin \gamma }\\ &=\displaystyle \frac{\cos \gamma }{\sin \gamma }+\displaystyle \frac{\sin ^{2}\gamma }{\sin \gamma }\\ &=\color{red}\cot \gamma +\sin \gamma \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 59.&\textrm{Jika diketahui}\: \: \sin \theta \cos \theta =\displaystyle \frac{3}{8},\\ &\textrm{maka nilai}\: \: \displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta }\: \: \textrm{adalah}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\displaystyle \frac{1}{4}\\ \textrm{b}.&\displaystyle \frac{1}{2}\\ \textrm{c}.&\displaystyle \frac{3}{4} \\ \textrm{d}.&\color{red}\displaystyle \frac{4}{3}\\ \textrm{e}.&\displaystyle 4 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta }\\ &=\sqrt{\left ( \displaystyle \frac{1}{\sin \theta }-\displaystyle \frac{1}{\cos \theta } \right )^{2}}\\ &=\sqrt{\displaystyle \frac{1}{\sin ^{2}\theta }-\displaystyle \frac{2}{\sin \theta \cos \theta }+\displaystyle \frac{1}{\cos ^{2}\theta }}\\ &=\sqrt{\displaystyle \frac{\cos ^{2}\theta +\sin ^{2}\theta }{\sin ^{2}\theta \cos ^{2}\theta }-\displaystyle \frac{2}{\sin \theta \cos \theta }}\\ &=\sqrt{\displaystyle \frac{1}{\left ( \displaystyle \frac{3}{8} \right )^{2}}-\displaystyle \frac{2}{\left ( \displaystyle \frac{3}{8} \right )}}=\sqrt{\displaystyle \frac{1}{\displaystyle \frac{9}{64}}-\displaystyle \frac{2}{\displaystyle \frac{3}{8}}}=\sqrt{\displaystyle \frac{64}{9}-\displaystyle \frac{16}{3}}\\ &=\sqrt{\displaystyle \frac{64}{9}-\displaystyle \frac{48}{9}}=\sqrt{\displaystyle \frac{16}{9}}=\color{red}\displaystyle \frac{4}{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 60.&\textrm{Bentuk}\: \: \displaystyle \frac{\left ( \sin \alpha +\sin \beta \right )\left ( \sin \alpha -\sin \beta \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ & \textrm{ekivalen dengan}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\tan \alpha -\tan \beta \\ \textrm{b}.&\tan \alpha +\tan \beta \\ \textrm{c}.&\color{red}\tan ^{2}\alpha -\tan ^{2}\beta \\ \textrm{d}.&\tan ^{2}\alpha +\tan ^{2}\beta \\ \textrm{e}.&\tan ^{3}\alpha -\tan ^{3}\beta \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}&\displaystyle \frac{\left ( \sin \alpha +\sin \beta \right )\left ( \sin \alpha -\sin \beta \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha -\sin ^{2}\beta }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha -\sin ^{2}\alpha \sin ^{2}\beta -\sin ^{2}\beta +\sin ^{2}\alpha \sin ^{2}\beta}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha\left ( 1-\sin ^{2}\beta \right ) -\sin ^{2}\beta \left ( 1-\sin ^{2}\alpha \right )}{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha \cos ^{2}\beta -\sin ^{2}\beta \cos ^{2}\alpha }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha \cos ^{2}\beta }{\left ( \cos \alpha \cos \beta \right )^{2}}-\displaystyle \frac{\sin ^{2}\beta \cos ^{2}\alpha }{\left ( \cos \alpha \cos \beta \right )^{2}}\\ &=\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\beta }{\cos ^{2}\beta }\\ &=\color{red}\tan ^{2}\alpha -\tan ^{2}\beta \end{aligned} \end{array}$.

DAFTAR PUSTAKA

1. Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
2. Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA