Contoh Soal 2 Induksi Matematika (Matematika Wajib Kelas XI)

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

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

$\begin{array}{l}\\ 8.& \text{Bila kita hendak membuktikan}{\displaystyle \sum _{i=1}^n={\displaystyle \frac{1}{2}n(n+1)}}\\ & \text{dengan induksi matematika}\\ & \text{maka untuk langkah}n=k+1\\ & \text{bentuk yang harus ditunjukkan adalah}...\\ & \begin{array}{l}\\ \text{a}.& 1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ \text{b}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ \text{c}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+2)}\\ \text{d}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+1)(k+2)}\\ \text{e}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+2)(k+3)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ P(k)& =1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ P(k+1)& =1+2+3+\cdots +k+(k+1)\\ & =\underset{{\displaystyle \frac{1}{2}k(k+1)}}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ & ={\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left({\displaystyle \frac{1}{2}k+1}\right)}\\ & =(k+1){\displaystyle \frac{1}{2}(k+2)}\\ & ={\displaystyle \frac{1}{2}(k+1)(k+2)}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}P(n)={\displaystyle \frac{n-1}{n+3},\text{maka}P(k+1)}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k-1}{k+3}}\\ \text{b}.& {\displaystyle \frac{k-1}{k+4}}\\ \text{c}.& {\displaystyle \frac{k}{k+4}}\\ \text{d}.& {\displaystyle \frac{k+1}{k+4}}\\ \text{e}.& {\displaystyle \frac{k+1}{k+5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)=& {\displaystyle \frac{n-1}{n+3}}\\ P(k+1)& ={\displaystyle \frac{k+1-1}{k+1+3}}\\ & ={\displaystyle \frac{k}{k+4}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}P(n)={\displaystyle \frac{n^2+1}{4},\text{maka}}\\ & \text{pernyataan untuk}P(k+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k^2+2k+1}{4}}\\ \text{b}.& {\displaystyle \frac{k^2+2k+2}{4}}\\ \text{c}.& {\displaystyle \frac{k^2+2k+2}{5}}\\ \text{d}.& {\displaystyle \frac{k^2+2k+3}{5}}\\ \text{e}.& {\displaystyle \frac{k^2+2k+3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}P(n)& ={\displaystyle \frac{n^2+1}{4}}\\ P(k+1)& ={\displaystyle \frac{(k+1{)}^2+1}{4}}\\ & ={\displaystyle \frac{k^2+2k+2}{4}}\end{array}\end{array}$

Contoh Soal 1 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \sum _{i=1}^{6}16i\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 306\\ \text{b}.& 314\\ \text{c}.& 326\\ \text{d}.& 336\\ \text{e}.& 402\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=1}^{6}16i}& =16.1+16.2+16.3+16.4+16.5+16.6\\ & =16+32+48+64+80+96\\ & =336\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Hasil dari}{\displaystyle \sum _{i=2}^9{i}^2\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 274\\ \text{b}.& 278\\ \text{c}.& 280\\ \text{d}.& 284\\ \text{e}.& 286\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=2}^9{i}^2}& =2^2+3^2+4^2+5^2+..+9^2\\ & =4+9+16+25+...+81\\ & =284\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Poa bilangan}12,14,16,18,20,...,(2n+10).\\ & \text{Nilai suku ke-100 adalah}....\\ & \begin{array}{l}\\ \text{a}.& 180\\ \text{b}.& 194\\ \text{c}.& 198\\ \text{d}.& 208\\ \text{e}.& 210\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{U}_n& =2n+10\\ U_{100}& =2\times 100+10\\ & =210\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa jika}\\ & 31+39+47+\cdots +8n+23=4n^2+27n\\ & \text{dengan}k,n\in \mathbb{N}\text{maka}\\ & 31+39+47+\cdots +8n+23+8k+31=....\\ & \begin{array}{l}\\ \text{a}.& 4k^2+27k\\ \text{b}.& 4k^2+35k\\ \text{c}.& 4k^2+35k+31\\ \text{d}.& 4k^2+35k+1\\ \text{e}.& 4k^2+35k+54\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \underset{4k^2+27k}{\underbrace{31+39+47+\cdots +8k+23}}+8k+31\\ & =4k^2+27k+8k+31\\ & =4k^2+35k+31\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan bahwa}n(n+1)(n+2)\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 4\\ \text{b}.& 5\\ \text{c}.& 6\\ \text{d}.& 7\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)& =n(n+1)(n+2)\\ P(1)& =1(1+1)(1+2)=1.2.3\\ & \text{adalah bilangan yang habis dibagi 6}\end{array}\end{array}$

Contoh Soal 6 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 26.& \text{Jika nilai}\cot A+\cos A=x\\ & \text{dan}\cot A-\cos A=y,\\ & \text{maka nilai}(x^2-y^2)=....\\ & \begin{array}{l}\\ \text{a}.& 4\sqrt{xy}\\ \text{b}.& 2\sqrt{xy}\\ \text{c}.& xy\\ \text{d}.& 2\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}xy& =(\cot A+\cos A)(\cot A-\cos A)\\ & ={\cot }^{2}A-{\cos }^{2}A\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}-{\cos }^{2}A\times \frac{{\sin }^{2}A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{{\cos }^{2}A}{{\sin }^{2}A}(1-{\sin }^{2}A)}\\ & ={\displaystyle \frac{{\cos }^{4}A}{{\sin }^{2}A}}\\ \sqrt{xy}& ={\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ \text{Se}& \text{lanjutnya}\\ x^2-y^2& ={(\cot A+\cos A)}^2-{(\cot A-\cos A)}^2\\ (x+y)& (x-y)=(\cot A+\cos A+\cot A-\cos A)\\ & \times (\cot A+\cos A-(\cot A-\cos A))\\ x^2-y^2& =2\cot A\times 2\cos A\\ & =4\times {\displaystyle \frac{\cos A}{\sin A}\times \cos A}\\ & =4\sqrt{xy}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& \text{Jika nilai}\cos A+\sin A=\sqrt{2}\cos A\\ & \text{maka nilai}(\cos A-\sin A)=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\cos A\\ \text{b}.& -\sqrt{2}\sin A\\ \text{c}.& \sqrt{2}\sin A\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}\sec A}\\ \text{e}.& {\displaystyle \frac{1}{\sqrt{2}}\csc A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\cos A+\sin A& =\sqrt{2}\cos A\\ {(\cos A+\sin A)}^2& ={(\sqrt{2}\cos A)}^2\\ 1+2\sin A\cos A& =2{\cos }^{2}A\\ 2\sin A\cos A& =2{\cos }^{2}A-1\\ \text{maka}& \\ {(\cos A-\sin A)}^2& =1-2\sin A\cos A\\ & =1-(2{\cos }^{2}A-1)\\ & =2-2{\cos }^{2}A\\ & =2(1-{\cos }^{2}A)\\ & =2{\sin }^{2}A\\ \cos A-\sin A& =\sqrt{2{\sin }^{2}A}\\ & =\sin A\sqrt{2}\\ & =\sqrt{2}.\sin A\end{array}\end{array}$

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

Contoh Soal 5 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 21.& \text{Nilai dari}{\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}=....}\\ & \begin{array}{l}\\ \text{a}.& -\mathrm{\infty }\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ e.& {\displaystyle \mathrm{\infty }}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{1}{{\sec }^{2}A}+\frac{1}{{\csc }^{2}A}}\\ & ={\cos }^{2}A+{\sin }^2=1\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Nilai dari}{\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}=....}\\ & \begin{array}{l}\\ \text{a}.& \cot B\times \cot C\\ \text{b}.& {\displaystyle \tan B\times \tan C}\\ \text{c}.& {\displaystyle \sec B\times \csc C}\\ \text{d}.& {\displaystyle \tan B\times \cot C}\\ e.& {\displaystyle \tan B\times \csc C}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan B+\tan C}{\cot B+\cot C}}\\ & ={\displaystyle \frac{\tan B+\tan C}{{\displaystyle \frac{1}{\tan B}+\frac{1}{\tan C}}}}\\ & ={\displaystyle \frac{\tan B+\tan C}{\left({\displaystyle \frac{\tan B+\tan C}{\tan B\times \tan C}}\right)}}\\ & =\tan B\times \tan C\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Nilai dari}\\ & {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=....}\\ & \begin{array}{l}\\ \text{a}.& 2\tan A\\ \text{b}.& 2\cot A\\ \text{c}.& {\displaystyle 2\sec A}\\ \text{d}.& {\displaystyle 2\csc A}\\ e.& {\displaystyle 2\tan A.\sec A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& {\displaystyle \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}}\\ & =\tan A\left({\displaystyle \frac{1}{{\displaystyle \frac{1}{\cos A}-1}}+\frac{1}{{\displaystyle \frac{1}{\cos A}+1}}}\right)\\ & ={\displaystyle \frac{\sin A}{\cos A}\left({\displaystyle \frac{\cos A}{1-\cos A}+\frac{\cos A}{1+\cos A}}\right)}\\ & ={\displaystyle \frac{\sin A}{1-\cos A}+\frac{\sin A}{1+\cos A}}\\ & ={\displaystyle \frac{\sin A(1+\cos A)+\sin A(1-\cos A)}{(1-\cos A)(1+\cos A)}}\\ & ={\displaystyle \frac{2\sin A}{1-{\cos }^2}}\\ & ={\displaystyle \frac{2\sin A}{{\sin }^{2}A}}\\ & ={\displaystyle \frac{2}{\sin A}}\\ & =2\csc A\end{array}\\ \\ & \text{Sebagai catatanya}\\ & \text{Anda bisa gunakan cara yang lain}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \sin (2x-{20}^{\circ })=-\cos (3x+{50}^{\circ })\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -{30}^{\circ }\\ \text{b}.& -{25}^{\circ }\\ \text{c}.& {20}^{\circ }\\ \text{d}.& {25}^{\circ }\\ e.& {30}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sin (2x-{20}^{\circ })& =-\cos (3x+{50}^{\circ })\\ \sin ({20}^{\circ }-2x)& =\cos (3x+{50}^{\circ })\\ \sin A& =\cos B,\text{artinya}\\ A+B& ={90}^{\circ },\text{maka}\\ ({20}^{\circ }-2x)+(3x+{50}^{\circ })& ={90}^{\circ }\\ x+{70}^{\circ }& ={90}^{\circ }\\ x& ={90}^{\circ }-{70}^{\circ }\\ & ={20}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & \tan (2x+{60}^{\circ })=\cot ({90}^{\circ }-3x)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {20}^{\circ }\\ \text{b}.& {30}^{\circ }\\ \text{c}.& {40}^{\circ }\\ \text{d}.& {50}^{\circ }\\ \text{e}.& {60}^{\circ }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\tan (2x+{60}^{\circ })& =\cot ({90}^{\circ }-3x)\\ \tan (2x+{60}^{\circ })& =\tan 3x\\ 2x+{60}^{\circ }& =3x\\ 2x-3x& =-{60}^{\circ }\\ -x& =-{60}^{\circ }\\ x& ={60}^{\circ }\end{array}\end{array}$

Contoh Soal 4 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 16.& \text{Jika}{\tan }^{2}x+\sec x=5\text{untuk}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\cos x=....}\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& {\displaystyle \frac{1}{2}}\\ \text{c}.& {\displaystyle \frac{1}{3}}\\ \text{d}.& {\displaystyle \frac{1}{\sqrt{2}}}\\ \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Ingat bahwa}& 0\le x\le {\displaystyle \frac{\pi }{2}}\\ \text{berarti sudu}& \text{t}x\text{berada di kuadran I}\\ \text{sehingga ak}& \text{an menyebabkan nilai}\\ & \cos x=+\\ \text{Selanjutnya}& \\ {\tan }^{2}x+\sec x& =5\\ {\sec }^{2}x-1+\sec x& =5\\ {\sec }^{2}x+\sec x-6& =0\\ (\sec x+3)(\sec x-2)& =0\\ \sec x=-3\text{atau}& \sec x=2\\ \text{untuk}\sec x& =-3(\text{tidak memenuhi})\\ \text{untuk}\sec x& =2(\text{memenuhi})\\ \text{Selanjutnya}& \text{lagi}\\ \sec x& =2\\ {\displaystyle \frac{1}{\cos x}}& =2\\ \cos x& ={\displaystyle \frac{1}{2}}\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Nilai}\\ & {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& {\displaystyle 2}\\ \text{c}.& {\displaystyle 3}\\ \text{d}.& {\displaystyle 3\cos A}\\ e.& {\displaystyle 4\sin A}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& {(\sin A+\cos A)}^2+{(\sin A-\cos A)}^2\\ & ={\sin }^{2}A+2\sin A\cos A+{\cos }^{2}A\\ & +{\sin }^{2}A-2\sin A\cos A+{\cos }^{2}A\\ & =1+1\\ & =2\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Nilai}\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}=....\\ & \begin{array}{l}\\ \text{a}.& \sec A+\tan A\\ \text{b}.& {\sec }^{2}A+{\tan }^{2}A\\ \text{c}.& {\sec }^{2}A-{\tan }^{2}A\\ \text{d}.& {\tan }^{2}A-{\sec }^{2}A\\ e.& \sec A\times \tan A\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}}}\\ & =\sqrt{{\displaystyle \frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{1-{\sin }^{2}A}}}\\ & =\sqrt{{\displaystyle \frac{{(1+\sin A)}^2}{{\cos }^{2}A}}}\\ & ={\displaystyle \frac{1+\sin A}{\cos A}}\\ & ={\displaystyle \frac{1}{\cos A}+\frac{\sin A}{\cos A}}\\ & =\sec A+\tan A\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Jika}{0}^{\circ }\le \theta ⩽{90}^{\circ },\text{maka nilai}\\ & \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)=....\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& {\displaystyle 0}\\ \text{c}.& {\displaystyle \frac{1}{4}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ e.& {\displaystyle 1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& \left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }-\frac{3+5\sin \theta }{4+5\cos \theta }}\right)\\ & =\left({\displaystyle \frac{5\cos \theta -4}{3-5\sin \theta }\times \frac{4+5\cos \theta }{4+5\cos \theta }}\right)\\ & -\left({\displaystyle \frac{3+5\sin \theta }{4+5\cos \theta }\times \frac{3-5\sin \theta }{3-5\sin \theta }}\right)\\ & ={\displaystyle \frac{25{\cos }^2-16-(9-25{\sin }^2\theta )}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{25({\sin }^2\theta +{\cos }^2\theta )-25}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & ={\displaystyle \frac{0}{(3-5\sin \theta )(4+5\cos \theta )}}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Nilai}\\ & (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )=....\\ & \begin{array}{l}\\ \text{a}.& -2\\ \text{b}.& {\displaystyle -1}\\ \text{c}.& {\displaystyle 0}\\ \text{d}.& {\displaystyle 1}\\ \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& (1+\cot \theta -\csc \theta )(1+\tan \theta +\sec \theta )\\ & =(1+{\displaystyle \frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }})(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })\\ & =\left({\displaystyle \frac{\sin \theta +\cos \theta -1}{\sin \theta }}\right)\left({\displaystyle \frac{\cos \theta +\sin \theta +1}{\cos \theta }}\right)\\ & ={\displaystyle \frac{{(\sin \theta +\cos \theta )}^2-1}{\sin \theta \cos \theta }}\\ & ={\displaystyle \frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }}\\ & =2\end{array}\end{array}$

Contoh Soal 3 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 11.& \text{Himpunan penyelesaian persamaan}\\ & 3\cos 2x+5\sin x+1=0\text{untuk}{0}^{\circ }\le x\le 2\pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{7}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{b}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{11}{6}\pi }\right\}\\ \text{c}.& \left\{{\displaystyle \frac{1}{6}\pi ,\frac{7}{6}\pi }\right\}\\ \text{d}.& \left\{{\displaystyle \frac{1}{5}\pi ,\frac{5}{6}\pi }\right\}\\ \text{e}.& \left\{{\displaystyle \frac{5}{6}\pi ,\frac{7}{6}\pi }\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}3\cos 2x+5\sin x+1& =0\\ 3(1-2{\sin }^{2}x)+5\sin x+1& =0\\ -6{\sin }^2+5\sin x+4& =0\\ 6{\sin }^{2}x-5\sin x-4& =0\\ (3\sin x-4)(2\sin x+1)& =0\\ \sin x={\displaystyle \frac{4}{3}\text{atau}\sin x}& =-\frac{1}{2}\\ \sin x& =\sin {150}^{\circ }=\frac{5}{6}\pi \\ x& =\{\begin{array}{ll}{\displaystyle \frac{7}{6}\pi }& +k.2\pi \\ \pi -{\displaystyle \frac{7}{6}\pi }& +k.2\pi \end{array}\\ \text{saat}k& =0\\ x_1& ={\displaystyle \frac{7}{6}\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi }\\ \text{saat}k& =1\\ x_1& ={\displaystyle \frac{7}{6}\pi +2\pi }\\ x_2& =-{\displaystyle \frac{1}{6}\pi +2\pi =\frac{11}{6}\pi }\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Himpunan penyelesaian dari}\\ & \sqrt{3}\sin 2x+2{\cos }^{2}x=-1\text{untuk}\\ & 0^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{240}^{\circ },{300}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{60}^{\circ }\}\\ \text{c}.& \{{150}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{120}^{\circ },{300}^{\circ }\}\\ \text{e}.& \{{60}^{\circ },{150}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sqrt{3}\sin 2x+2{\cos }^{2}x& =-1\\ \sqrt{3}\sin 2x+1+\cos 2x& =-1\\ \sqrt{3}\sin 2x+\cos 2x& =-2\\ \sqrt{{\sqrt{3}}^2+1^2}\cos (2x-\alpha )& =-2\\ a=x=1,b& =y=\sqrt{3}\\ \alpha & =\arctan {\displaystyle \frac{b}{a}=\arctan \frac{\sqrt{3}}{1}}\\ \alpha & ={60}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (2x-{60}^{\circ })& =-2\\ \cos (2x-{60}^{\circ })& =-1\\ \cos (2x-{60}^{\circ })& =\cos {180}^{\circ }\\ (2x-{60}^{\circ })& =\pm {180}^{\circ }+k{.360}^{\circ }\\ 2x& ={60}^{\circ }\pm {180}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {90}^{\circ }+k{.180}^{\circ }\\ \text{saat}k& =0\\ x_1& ={120}^{\circ }\\ x_2& =-{60}^{\circ }\\ \text{saat}k& =1\\ x_3& ={120}^{\circ }+{360}^{\circ }=....\\ x_2& =-{60}^{\circ }+{360}^{\circ }={300}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Jika}3\sin \theta +4\cos \theta =5\text{maka}\\ & \text{nilai dari}\sin \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0,3\\ \text{b}.& 0,60\\ \text{c}.& 0,75\\ \text{d}.& 0,80\\ \text{e}.& 1,20\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}3\sin \theta +4\cos \theta & =5\\ k\cos (\theta -\alpha )& =5\\ a=x=4,& b=y=3\\ \theta & =\arctan {\displaystyle \frac{b}{a}}\\ \theta & =\arctan {\displaystyle \frac{3}{4}}\\ \text{atau}& \\ \tan \theta & ={\displaystyle \frac{3}{4}}\\ \text{maka}& \\ \sin \theta & ={\displaystyle \frac{3}{\sqrt{3^2+4^2}}}\\ & ={\displaystyle \frac{3}{5}}\\ & =0,6\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Jika}\tan \theta +\sec \theta =x\text{maka}\\ & \text{nilai dari}\tan \theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{2x}{x^2-1}}\\ \text{b}.& {\displaystyle \frac{2x}{x^2+1}}\\ \text{c}.& {\displaystyle \frac{x^2+1}{2x}}\\ \text{d}.& {\displaystyle \frac{x^2-1}{2x}}\\ \text{e}.& {\displaystyle \frac{x^2-1}{x^2+1}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\text{Langkah}& 1\\ \tan \theta +\sec \theta & =x\\ {\displaystyle \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }}& =x\\ {\displaystyle \frac{\sin \theta +1}{\cos \theta }}& =x\\ \sin \theta +1& =x\cos \theta ...........1\\ \text{Langkah}& 2\\ {(\tan \theta +\sec \theta )}^2& =x^2\\ {\tan }^2\theta +2\tan \theta \sec \theta +{\sec }^2\theta & =x^2\\ {\sec }^2\theta -1+2\tan \theta \sec \theta & +{\sec }^2\theta =x^2\\ 2{\sec }^2\theta +2\tan \theta \sec \theta & =x^2+1\\ {\displaystyle \frac{2}{{\cos }^2\theta }+\frac{2\sin \theta }{{\cos }^2\theta }}& =x^2+1\\ 1+\sin \theta & =\left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2\theta .....2\\ \text{Langkah}& 3\\ \left({\displaystyle \frac{x^2+1}{2}}\right){\cos }^2& =x\cos \theta \\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \text{maka}& (\mathbf{\text{dengan sisi segitiga}})\\ \tan \theta ={\displaystyle \frac{\sqrt{{(x^2+1)}^2-(2x{)}^2}}{2x}}& ={\displaystyle \frac{\sqrt{x^4+2x^2+1-4x^2}}{2x}}\\ \tan \theta & ={\displaystyle \frac{\sqrt{x^4-2x^2+1}}{2x}}\\ & ={\displaystyle \frac{\sqrt{{(x^2-1)}^2}}{2x}}\\ & ={\displaystyle \frac{x^2-1}{2x}}\end{array}\end{array}$

$\begin{array}{rl}.\mathbf{\text{Sehingga}}& \mathbf{\text{dari kasus di atas didapatkan}}\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1}}\\ \cos \theta & ={\displaystyle \frac{2x}{x^2+1}}\\ \tan \theta & ={\displaystyle \frac{x^2-1}{2x}}\end{array}$

$\begin{array}{l}\\ 15.& \text{Jika}\sec x+\tan x={\displaystyle \frac{3}{2}\text{untuk}}\\ & 0\le x\le {\displaystyle \frac{\pi }{2}\text{maka nilai}\sin x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{5}{13}}\\ \text{b}.& {\displaystyle \frac{12}{13}}\\ \text{c}.& {\displaystyle 1}\\ \text{d}.& {\displaystyle \frac{2}{13}}\\ \text{e}.& {\displaystyle \frac{5}{12}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\sec x+\tan x& ={\displaystyle \frac{3}{2}}\\ \text{ingat saat}& \text{mengerjakan soal no}.14\\ \sec \theta +\tan \theta & =x\\ \sin \theta & ={\displaystyle \frac{x^2-1}{x^2+1},\text{maka}}\\ \sin x& ={\displaystyle \frac{{\left({\displaystyle \frac{3}{2}}\right)}^2-1}{{\left({\displaystyle \frac{3}{2}}\right)}^2+1}}\\ & ={\displaystyle \frac{{\displaystyle \frac{9}{4}-1}}{{\displaystyle \frac{9}{4}+1}}}\\ & =\frac{{\displaystyle \frac{5}{4}}}{{\displaystyle \frac{13}{4}}}\\ & ={\displaystyle \frac{5}{13}}\end{array}\end{array}$

Contoh Soal 2 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 6.& \text{Bentuk}\sqrt{3}\cos x-\sin x\text{untuk}0\le x\le 2\pi ,\\ & \text{dapat dinyatakan sebagai}....\\ & \begin{array}{l}\\ \text{a}.& 2\cos (x+{\displaystyle \frac{\pi }{6}})\\ \text{b}.& 2\cos (x+{\displaystyle \frac{7\pi }{6}})\\ \text{c}.& 2\cos (x-{\displaystyle \frac{11\pi }{6}})\\ \text{d}.& 2\cos (x-{\displaystyle \frac{7\pi }{6}})\\ \text{e}.& 2\cos (x-{\displaystyle \frac{\pi }{6}})\end{array}\\ \\ & \text{Jawab}:\text{c}\\ & \begin{array}{rl}\sqrt{3}\cos x-\sin x& =k\cos (x-\alpha )\\ (1)& (a,b)=\{\begin{array}{ll}a& =\sqrt{3}\\ b& =-1\end{array}\\ \text{maka}& \text{titik ada dikadran IV}\\ (2)k& =\sqrt{a^2+b^2}\\ & =\sqrt{{\sqrt{3}}^2+(-1{)}^2}\\ & =\sqrt{4}=2\\ (3)\alpha & =\arctan \frac{b}{a}=\arctan \left(\frac{-1}{\sqrt{3}}\right)=-{30}^{\circ }\\ & =({360}^{\circ }-{30}^{\circ })={330}^{\circ }={\displaystyle \frac{11}{6}\pi }\\ \text{sehingga}& \\ \sqrt{3}\cos x-\sin x& =2\cos (x-{\displaystyle \frac{11}{6}\pi })\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Nilai-nilai}x\text{yang terletak pada}0\le x\le 2\pi ,\\ & \text{yang memenuhi persamaan}\sqrt{3}\cos x+\sin x=\sqrt{2}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {75}^{\circ }\text{atau}{285}^{\circ }\\ \text{b}.& {75}^{\circ }\text{atau}{345}^{\circ }\\ \text{c}.& {15}^{\circ }\text{atau}{285}^{\circ }\\ \text{d}.& {15}^{\circ }\text{atau}{345}^{\circ }\\ \text{e}.& {15}^{\circ }\text{atau}{75}^{\circ }\end{array}\\ \\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}\sqrt{3}\cos x+\sin x& =\sqrt{2}\\ \sqrt{{\sqrt{3}}^2+1^2}& (\cos (\alpha -\arctan {\displaystyle \frac{1}{\sqrt{3}}}))=\sqrt{2}\\ 2\cos (x-{30}^{\circ })& =\sqrt{2}\\ \cos (x-{30}^{\circ })& ={\displaystyle \frac{\sqrt{2}}{2}=\frac{1}{2}\sqrt{2}}\\ \cos (x-{30}^{\circ })& =\cos {45}^{\circ }\\ (x-{30}^{\circ })& =\pm {45}^{\circ }+k{.360}^{\circ }\\ x& ={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ \text{saat}k& =0\\ x_1& ={75}^{\circ }\\ x_2& =-{15}^{\circ }\\ \text{saat}k& =1\\ x_3& ={75}^{\circ }+{360}^{\circ }={435}^{\circ }\\ x_4& =-{15}^{\circ }+{360}^{\circ }={345}^{\circ }\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Diketahui fungsi trigonometri}f(x)={\displaystyle \frac{1}{2}\sin 3x}\\ & \text{perhatikanlah pernyataan-pernyataan berikut}\\ & (1)\text{hasil dari}f(0)+f\left({\displaystyle \frac{\pi }{6}}\right)=1\\ & (2)\text{hasil dari}f\left({\displaystyle \frac{\pi }{6}}\right)+f\left({\displaystyle \frac{\pi }{3}}\right)={\displaystyle \frac{1}{2}}\\ & (3)\text{hasil dari}f\left({\displaystyle \frac{\pi }{6}}\right)-f\left({\displaystyle \frac{\pi }{3}}\right)={\displaystyle \frac{1}{2}}\\ & (4)\text{hasil dari}f\left({\displaystyle \frac{\pi }{3}}\right)-f\left({\displaystyle \frac{\pi }{6}}\right)=1\\ & \text{Pernyataan yang tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& (1)\text{dan}(2)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (1)\text{dan}(4)\\ \text{d}.& (2)\text{dan}(3)\\ \text{e}.& (3)\text{dan}(4)\end{array}\\ \\ & \text{Jawab}:\text{d}\\ & \begin{array}{rl}\text{Diketahui}& \\ f(x)& ={\displaystyle \frac{1}{2}\sin 3x}\\ \text{maka}& \\ f(0)& ={\displaystyle \frac{1}{2}\sin 3(0^{\circ })=0}\\ f\left({\displaystyle \frac{\pi }{3}}\right)& ={\displaystyle \frac{1}{2}\sin 3\left({\displaystyle \frac{\pi }{3}}\right)=0}\\ f\left({\displaystyle \frac{\pi }{6}}\right)& ={\displaystyle \frac{1}{2}\sin 3\left({\displaystyle \frac{\pi }{6}}\right)={\displaystyle \frac{1}{2}}}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Himpunan penyelesaian dari}\sin x-\sqrt{3}\cos x=-1\\ & \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{0^{\circ },{120}^{\circ }\}\\ \text{b}.& \{{90}^{\circ },{330}^{\circ }\}\\ \text{c}.& \{{60}^{\circ },{180}^{\circ }\}\\ \text{d}.& \{{90}^{\circ },{120}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}\sin x-\sqrt{3}\cos x& =-1\\ \sqrt{1^2+{(-\sqrt{3})}^2}\cos (x-\alpha )& =-1\\ a& =x=-\sqrt{3},b=y=1\\ & \mathbf{\text{kuadran II}}\\ \alpha & =\arctan \left({\displaystyle \frac{1}{-\sqrt{3}}}\right)=-{30}^{\circ }\\ & ={180}^{\circ }-{30}^{\circ }={150}^{\circ }\\ \text{maka persamaan akan}& \text{menjadi}\\ 2\cos (x-{150}^{\circ })& =-1\\ \cos (x-{150}^{\circ })& ={\displaystyle \frac{-1}{2}}\\ \cos (x-{150}^{\circ })& =\cos {120}^{\circ }\\ (x-{150}^{\circ })& =\pm {120}^{\circ }+k{.360}^{\circ }\\ x& ={150}^{\circ }\pm {120}^{\circ }+k{.360}^{\circ }\\ \text{saat}& k=0\\ x_1& ={270}^{\circ }\\ x_2& ={30}^{\circ }\\ \text{saat}& k=1\\ x_3& ={270}^{\circ }+{360}^{\circ }=....\\ x_4& ={30}^{\circ }+{360}^{\circ }=....\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Nilai}\tan x\text{yang memenuhi persamaan}\\ & \cos 2x+7\cos x-3=0\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \sqrt{3}\\ \text{b}.& {\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& {\displaystyle \frac{1}{3}\sqrt{3}}\\ \text{d}.& {\displaystyle \frac{1}{2}}\\ \text{e}.& {\displaystyle \frac{1}{5}\sqrt{5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\cos 2x+7\cos x-3& =0\\ 2{\cos }^{2}x-1+7\cos x-3& =0\\ 2{\cos }^{2}x+7\cos x-4& =0\\ (\cos x+4)(2\cos -1)& =0\\ \cos x=-4\text{atau}\cos x& ={\displaystyle \frac{1}{2}=\cos {60}^{\circ }}\\ x& ={60}^{\circ },\\ \text{maka}& \\ \tan {60}^{\circ }& =\sqrt{3}\\ \\ \text{ dan ingat bahwa}& \cos x=-4\text{tidak memenuhi}\end{array}\end{array}$

Contoh Soal 1 Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\begin{array}{l}\\ 1.& \text{Himpunan penyelesaian dari}\\ & \sin 2x={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{30}^{\circ },{210}^{\circ }\}\\ \text{b}.& \{{60}^{\circ },{240}^{\circ }\}\\ \text{c}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ }\}\\ \text{d}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ }\}\\ \text{e}.& \{{30}^{\circ },{60}^{\circ },{210}^{\circ },{240}^{\circ },{270}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\sin 2x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \sin 2x& =\sin {60}^{\circ }\\ 2x& =\{\begin{array}{ll}{60}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{60}^{\circ })& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.180}^{\circ }\\ {60}^{\circ }& +k{.180}^{\circ }\end{array}\\ \text{saat}& k=0\\ x& =\{\begin{array}{ll}{30}^{\circ }& \\ {60}^{\circ }& \end{array}\\ \text{saat}& k=1\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{1.180}^{\circ }={210}^{\circ }\\ {60}^{\circ }& +{1.180}^{\circ }={240}^{\circ }\end{array}\\ \text{saat}& k=2\\ x& =\{\begin{array}{ll}{30}^{\circ }& +{2.180}^{\circ }={390}^{\circ }\\ {60}^{\circ }& +{2.180}^{\circ }={420}^{\circ }\end{array}\\ \text{kedua}& \text{nnya tidak memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Himpunan penyelesaian dari}\\ & \tan 2x-\sqrt{3}=0\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{b}.& \{{30}^{\circ },{120}^{\circ },{210}^{\circ },{300}^{\circ }\}\\ \text{c}.& \{{45}^{\circ },{135}^{\circ },{225}^{\circ },{315}^{\circ }\}\\ \text{d}.& \{{15}^{\circ },{105}^{\circ },{195}^{\circ },{285}^{\circ }\}\\ \text{e}.& \{{15}^{\circ },{30}^{\circ },{45}^{\circ },{60}^{\circ },{75}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\tan 2x& -\sqrt{3}=0\\ \tan 2x& =\sqrt{3}\\ \tan 2x& =\tan {60}^{\circ }\\ 2x& =60+k{.180}^{\circ }\\ x& ={30}^{\circ }+k{.90}^{\circ }\\ \text{saat}& k=0\\ x& ={30}^{\circ }\\ \text{saat}& k=1\\ x& ={30}^{\circ }+{90}^{\circ }={120}^{\circ }\\ \text{saat}& k=2\\ x& ={30}^{\circ }+{180}^{\circ }={210}^{\circ }\\ \text{saat}& k=3\\ x& ={30}^{\circ }+{270}^{\circ }={300}^{\circ }\\ \text{saat}& k=4\\ x& ={30}^{\circ }+{360}^{\circ }={390}^{\circ }\\ \text{tidak}& \text{memenuhi}\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Himpunan penyelesaian dari}\\ & \cos 3x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }\le x\le {180}^{\circ }}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{{40}^{\circ },{80}^{\circ }\}\\ \text{b}.& \{{50}^{\circ },{70}^{\circ }\}\\ \text{c}.& \{{40}^{\circ },{70}^{\circ },{80}^{\circ }\}\\ \text{d}.& \{{50}^{\circ },{70}^{\circ },{170}^{\circ }\}\\ \text{e}.& \{{50}^{\circ },{80}^{\circ },{170}^{\circ }\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}\cos 3x& =-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos 3x& =-\cos {30}^{\circ }\\ \cos 3x& =\cos ({180}^{\circ }-{30}^{\circ })=\cos {150}^{\circ }\\ 3x& =\pm {150}^{\circ }+k{.360}^{\circ }\\ x& =\pm {50}^{\circ }+k{.120}^{\circ }\\ \text{saat}& k=0\\ x& =\pm {50}^{\circ }\to x={50}^{\circ }(\text{mm})\\ \text{saat}& k=1\\ x& =\pm {50}^{\circ }+{120}^{\circ }=\{\begin{array}{ll}{170}^{\circ }& (\text{mm})\\ {70}^{\circ }& (\text{mm})\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Nilai}x\text{yang memenuhi persamaan}\\ & 2{\cos }^{2}x+\cos x-1=0\text{untuk}0\le x\le \pi \\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{3}\pi \text{dan}\pi }\\ \text{b}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{2}{3}\pi }\\ \text{c}.& {\displaystyle \frac{1}{3}\pi \text{dan}\frac{3}{4}\pi }\\ \text{d}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{3}{4}\pi }\\ \text{e}.& {\displaystyle \frac{1}{4}\pi \text{dan}\frac{2}{3}\pi }\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}2{\cos }^{2}x+\cos x-1& =0\\ (2\cos x-1)(\cos x+1)& =0\\ \cos x={\displaystyle \frac{1}{2}\text{atau}}& \cos x=-1\\ \cos x=\cos {60}^{\circ }=\cos \frac{1}{3}\pi & \\ \text{atau}\cos x& =\cos {180}^{\circ }=\cos \pi \end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Untuk}x\text{yang memenuhi persamaan}\\ & {\tan }^{2}x-\tan x-6=0\text{pada}0\le x\le \pi ,\\ & \text{maka himpunan nilai}\sin x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\\ \text{b}.& \left\{{\displaystyle \frac{3\sqrt{10}}{10},-\frac{2\sqrt{5}}{5}}\right\}\\ \text{c}.& \{-{\displaystyle \frac{3\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\}\\ \text{d}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{\sqrt{5}}{5}}\right\}\\ \text{e}.& \left\{{\displaystyle \frac{\sqrt{10}}{10},\frac{2\sqrt{5}}{5}}\right\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\tan }^{2}x-\tan x-6& =0\\ (\tan x-3)(\tan x+2)& =0\\ \tan x=3\text{atau}& \tan x=-2\\ \tan x={\displaystyle \frac{3}{1}\text{atau}}& \tan x=\frac{-2}{1}\\ \sin x={\displaystyle \frac{3}{\sqrt{1^2+3^2}}\text{atau}}& \sin x=\frac{2}{\sqrt{1^2+2^2}}\\ \sin x={\displaystyle \frac{3}{\sqrt{10}}\text{atau}}& \sin x=\frac{2}{\sqrt{5}}\\ \sin x={\displaystyle \frac{3}{10}\sqrt{10}\text{atau}}& \sin x=\frac{2}{5}\sqrt{5}\end{array}\end{array}$

Lanjutan 3 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{3. Grafik Fungsi Tangen}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 0& \frac{1}{3}\sqrt{3}& 1& \sqrt{3}& \mathrm{\infty }& -\sqrt{3}& -1& -\frac{1}{3}\sqrt{3}& 0\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& \frac{1}{3}\sqrt{3}& 1& \sqrt{3}& \mathrm{\infty }& -\sqrt{3}& -1& -\frac{1}{3}\sqrt{3}& 0& \\ \hline\end{array}$

Pada fungsi Tangen demikian juga nanti Cotangen ada beberapa nilai yang tida terdefinisi. Dalam fungsi Tangen fungsi, nilai fungsi yang tidak terdefini terdapat saat nilai  $x={\displaystyle \frac{\pi }{2}={90}^{\circ }}$ dan $x={\displaystyle \frac{3\pi }{2}={270}^{\circ }}$. Sehingga pada saat posisi nilai itu, maka dibuatlah garis putus-putus pada grafik yang dan ditampakkan berupa grais vertikal yang selanjutnya garis vertikal itu disebut sebagai asimtot.

LIMIT FUNGSI ALJABAR

 $\text{A. Pendahuluan}$

Mengingat kembali definisi limit yang telah dipelajari sebelumnya di kelas XI, yaitu limit fungsi aljabar $f(x)$ yang didefinisikan dengan:

$\begin{array}{rl}\underset{x\to a}{\text{lim}}f(x)=L& \mathbf{\text{adalah}}\text{Jika}x\text{mendekati}a\\ & \text{dengan tidak sama dengan}a,\\ & \text{maka nilai}f(x)\text{mendekati}L\end{array}$.

Perhatikan definisi di atas istilah  $x\text{mendekati}a$ dituliskan dengan simbol  $(x\to a)$. Suatu nilai limit dianggap ada jika nilai $f(x)$ mendekati  $a$ dari arah kiri sama dengan nilai $f(x)$ mendekati  $a$ dari arah kanan dengan nilai yang sama misalnya $L$. Jika disimbolkan pernyataan ini menjadi berikut

$\begin{array}{rl}\underset{x\to a^{-}}{\text{lim}}f(x)=\underset{x\to a^{+}}{\text{lim}}f(x)=\underset{x\to a}{\text{lim}}f(x)=L& \end{array}$.

$\begin{array}{rl}\text{Perlu di}& \text{perhatikan bahwa didekati dari}\\ \bullet \mathbf{\text{kiri}}& \text{disimbolkan dengan}\underset{x\to a^{-}}{\text{lim}}f(x),\text{dan}\\ \bullet \mathbf{\text{kan}}& \mathbf{\text{an}}\text{disimbolkan dengan}\underset{x\to a^{+}}{\text{lim}}f(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukanlah nilai limit dari}f(x)={\displaystyle \frac{x^2-4}{x-2}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika fungsi}{\displaystyle \frac{x^2-4}{x-2}}\\ & \text{di sekitar}x=2\text{sebagaimana dalam tabel}\\ & \text{berikut}\end{array}\end{array}$.

$\begin{array}{rl}.& \text{Jadi, nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}=2\text{atau dapat dikatakan}}\\ & \text{nilai}\underset{x\to 2}{\text{lim}}{\displaystyle \frac{x^2-4}{x-2}\mathbf{\text{ada}}}\\ & \text{meskipun nilai substitusi langsung}x=2\text{yaitu}\\ & f(0)={\displaystyle \frac{0^2-0}{0-0}=\frac{0}{0}\text{berupa bentuk tak}}\\ & \text{tentu. Berikut ilustrasinya}\end{array}$

$\begin{array}{l}\\ 2.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 5}{\text{lim}}f(x),\text{untuk}f(x)=\{\begin{array}{ll}x& \text{saat}x<5\\ 5-x& \text{saat}x\ge 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 5^{-},\text{maka}\underset{x\to 5^{-}}{\text{lim}}f(x)=x\\ & \text{atau}\underset{x\to 5^{-}}{\text{lim}}f(x)=5\\ & \text{boleh juga dituliskan dengan}\\ & \underset{x\to 5^{-}}{\text{lim}}f(x)=x=5.\text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 5^{+},\\ & \text{maka}\underset{x\to 5^{+}}{\text{lim}}f(x)=\underset{x\to 5^{+}}{\text{lim}}(5-x)=5-5=0.\\ & \text{Karena nilai}\underset{x\to 5^{-}}{\text{lim}}f(x)\ne \underset{x\to 5^{+}}{\text{lim}}f(x),\text{maka}\\ & \text{nilai atau harga}\underset{x\to 5}{\text{lim}}f(x)\mathbf{\text{tidak ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Selidikilah limit fungsi berikut apakah}\\ & \text{memiliki harga limit}\\ & \underset{x\to 0}{\text{lim}}f(x),\text{untuk}f(x)=\cos x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Perhatikanlah ketika didekati dari kiri}\\ & \text{yaitu}x\to 0^{-},\text{maka}\underset{x\to 0^{-}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& -0,5& -0,4& -0,3& -0,2& -0,1& 0\\ \cos x& ...& ...& 0,999986& 0,999994& 0,9999985& 1\\ \hline\end{array}\\ & \text{Sedangkan ketika}\\ & \text{didekati dari arah kanan yaitu}x\to 0^{+},\\ & \text{maka}\underset{x\to 0^{+}}{\text{lim}}\cos x\\ & \begin{array}{|ccccccc|}\hline x& 0& 0,1& 0,2& 0,3& 0,4& 0,5\\ \cos x& 1& 0,9999985& 0,999994& 0,999986& ...& ...\\ \hline\end{array}\\ & \text{Karena nilai}\underset{x\to 0^{-}}{\text{lim}}f(x)=\underset{x\to 0^{+}}{\text{lim}}f(x)=1,\text{maka}\\ & \text{nilai}\underset{x\to 0}{\text{lim}}\cos x\mathbf{\text{ ada}}\\ & \text{Berikut ilustrasi gambarnya}\end{array}\end{array}$.

$\text{B. Sifat-Sifat Limit Fungsi}$

$\begin{array}{rl}& \text{Misalkan}f\text{dan}g\text{adalah fungsi-fungsi yang}\\ & \text{mempunyai nilai limit di titik sekitar}x=a\\ & \text{atau}(x\to a)\text{dan}c\text{adalah suatu konstanta}\\ & \text{serta}n\text{adalah suatu bilangan bulat positif},\\ & \text{maka berlaku sifat-sifat berikut}:\\ & \begin{array}{l}\\ 1.& \underset{x\to a}{\text{lim}}{\displaystyle c=c}\\ 2.& \underset{x\to a}{\text{lim}}{\displaystyle x^n=a^n}\\ 3.& \underset{x\to a}{\text{lim}}c.f(x)=c.\underset{x\to a}{\text{lim}}f(x)\\ 4.& \underset{x\to a}{\text{lim}}(f(x)\pm g(x))=\underset{x\to a}{\text{lim}}f(x)\pm \underset{x\to a}{\text{lim}}g(x)\\ 5.& \underset{x\to a}{\text{lim}}(f(x)\times g(x))=\underset{x\to a}{\text{lim}}f(x)\times \underset{x\to a}{\text{lim}}g(x)\\ 6.& \underset{x\to a}{\text{lim}}{\displaystyle \frac{f(x)}{g(x)}={\displaystyle \frac{\underset{x\to a}{\text{lim}}f(x)}{\underset{x\to a}{\text{lim}}g(x)}}}\\ 7.& \underset{x\to a}{\text{lim}}{(f(x))}^n={[\underset{x\to a}{\text{lim}}f(x))]}^n\\ 8.& \underset{x\to a}{\text{lim}}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to \mathrm{\infty }}{\text{lim}}f(x)},\text{dengan}\underset{x\to a}{\text{lim}}f(x)\ge 0\\ & \text{dan}n\text{genap}\end{array}\end{array}$

Lanjutan 2 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{2. Grafik Fungsi Cosinus}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 1& \frac{1}{2}\sqrt{3}& \frac{1}{2}\sqrt{2}& \frac{1}{2}& 0& -\frac{1}{2}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}\sqrt{3}& -1\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& -\frac{1}{2}\sqrt{3}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}& 0& \frac{1}{2}& \frac{1}{2}\sqrt{2}& \frac{1}{2}\sqrt{3}& 1& \\ \hline\end{array}$

Notasi Sigma Lanjutan Induksi Matematika (Matematika Wajib Kelas XI)

$\text{A. Pendahuluan}$

Notasi sigma dari asalnya dari yaitu dari huruf yunani yang memiliki makna jumlah. Dalam matematika lambang notasi sigma $"\sum "$  selanjutnya akan menunjukkan penjumlahan yang teratur sehingga penulisan sebuah deret dari suatu bilangan yang berpola tertentu dapat disederhanakan lebih ringkas.

Sebagai ilustrasinya untuk deretny adalah sebagai berikut

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100\\ \text{b}.1+3+5+7+9+\cdots +199\\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2\end{array}$

Dari bentuk deret di atas jika dimodelkan dengan notasi sigma maka bentuknya akan menjadi lebih sederhana, yaitu:

$\begin{array}{rl}& \sum _{i=1}^n{a}_i=a_1+a_2+a_3+\cdots +a_n\\ & \text{Dibaca}:"\text{Jumlah}\text{dari}{a}_i\text{untuk}i\\ & \text{dari 1 sampai dengan}n"\text{dan}{a}_i\text{adalah suku ke}-i\end{array}$

Sehingga contoh ilustrasi deret di atas jika dinotasikan dengan notasi sigma menjadi

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100=\sum _{i=1}^{100}i\\ \\ \text{b}.1+3+5+7+9+\cdots +199=\sum _{i=1}^{100}(2i-1)\\ \\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2=\sum _{i=1}^{100}{i}^2\end{array}$

$\text{B. Sifat-Sifat Notasi Sigma}$

Misalkan diketahui $a_k$  dan $b_k$  adalah suku ke-k dan C adalah sebuah konstanta, maka

$\begin{array}{l}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C=nC}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C.a_k=C\sum _{k=1}^n{a}_k}\\ & ⧫{\displaystyle \sum _{k=1}^n(a_k+b_k)={\displaystyle \sum _{k=1}^n{a}_k+\sum _{k=1}^n{b}_k}}\\ & ⧫{\displaystyle \sum _{k=1}^n{(a_k+b_k)}^2={\displaystyle \sum _{k=1}^n{a}_k^2+2\sum _{k=1}^n{a}_k{b}_k+\sum _{k=1}^n{b}_k^2}}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^{n-1}{a}_k+a_n}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^m{a}_k+\sum _{k=m+1}^n{a}_k,1<m<n}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Uraikan jumlah berikut dengan lengkap}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k}& \text{f}.{\displaystyle \sum _{k=1}^3{2}^k}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)}& \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k}& \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)}& \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)}& \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)}\end{array}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & \begin{array}{l}\\ 1.& \text{Perhatikanlah,}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k=1+2+3+4=10}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k=5.1+5.2+5.3+5.4=50}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)=30}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)=(2.1+3)+(2.2+3)+(2.3+3)=21}\\ & \text{f}.{\displaystyle \sum _{k=1}^3{2}^k=2^1+2^2+2^3+2^4=2+4+8+16=30}\\ & \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}={\displaystyle \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}={\displaystyle \frac{9+3+1}{27}=\frac{13}{27}}}}\\ & \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)=\cdots +\cdots +\cdots +\cdots +\cdots }\\ & \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k=\cdots +\cdots +\cdots +\cdots }\\ & \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)=\cdots +\cdots +\cdots +\cdots +\cdots }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ & \text{a}.2+4+8+16+32+64\\ & \text{b}.2+6+18+54+162\\ & \text{c}.15+24+35+48\\ & \text{d}.{\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}}\\ & \text{e}.ab+a^2{b}^2+a^3{b}^3+a^4{b}^4\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a})2+4+8+16+32+64={\displaystyle \sum _{k=1}^6{2}^k}\\ & (\text{b})2+6+18+54+162={\displaystyle \sum _{k=1}^5{2.3}^{k-1}}\\ & (\text{c})15+24+35+48={\displaystyle \sum _{k=1}^4(k^2+6k+8)}\\ & (\text{d}){\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}={\displaystyle \sum _{k=1}^5\frac{2^k}{(2k+1)}}}\\ & (\text{e})ab+a^2{b}^2+a^3{b}^3+a^4{b}^4={\displaystyle \sum _{k=1}^4(ab{)}^k}\end{array}$

$\begin{array}{l}\\ 3.& \text{Dengan menggunakan kaidah notasi sigma},\\ & \text{tunjukkan bahwa}\\ & \text{a}.{\displaystyle \sum _{k=1}^6(2k+3)=2\sum _{k=1}^{6}k+18}\\ & \text{b}.{\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=1}^{6}k+30}\\ & \text{c}.{\displaystyle \sum _{k=2}^5(2k^2+3k+3)=2\sum _{k=1}^4{k}^2+7\sum _{k=1}^{4}k+32}\\ & \text{d}.{\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=1}^6{k}^2-2\sum _{k=1}^{6}k+6}\\ & \text{e}.{\displaystyle \sum _{k=3}^6(k^2+2k-3)=\sum _{k=1}^6{k}^2+6\sum _{k=1}^4+20}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a}){\displaystyle \sum _{k=1}^6(2k+3)=\sum _{k=1}^{6}2k+\sum _{k=1}^{6}3}\\ & =\sum _{k=1}^{6}2k+6.3=2\sum _{k=1}^{6}k+18\\ & (\text{b}){\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=3-2}^{8-2}((k+2)+3)}\\ & =\sum _{k=1}^6(k+5)=\sum _{k=1}^{6}k+\sum _{k=1}^{6}5=\sum _{k=1}^{6}k+6.5\\ & =\sum _{k=1}^{6}k+30\\ & (\text{c}){\displaystyle \sum _{k=2}^5(2k^2+3k+3)}\\ & =\sum _{k=2-1}^{5-1}(2(k+1{)}^2+3(k+1)+3)=\cdots \\ & (\text{d}){\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=0+1}^{5+1}(k-1{)}^2=\cdots }\\ & (\text{e}){\displaystyle \sum _{k=3}^6(k^2+2k-3)=\cdots }\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{l}\\ .& \text{Buktikanlah bahwa}\\ & \text{a}.{\displaystyle \sum _{k=6}^{12}{k}^2=\sum _{k=1}^7{k}^2+10\sum _{k=1}^{7}k+175}\\ & \text{b}.{\displaystyle \sum _{k=1}^n(3k-1{)}^2=9\sum _{k=1}^n{k}^2-6\sum _{k=1}^{n}k+n}\\ & \text{c}.{\displaystyle \sum _{k=m}^n{a}_k=\sum _{k=m+p}^{n+p}{a}_{k-p}}\\ & \text{d}.{\displaystyle \sum _{i=m}^n{a}_i=\sum _{i=1}^n{a}_i-\sum _{i=1}^{m-1}{a}_i}\\ & \text{e}.{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=0}^{n-1}{a}_{k+1}=\sum _{k=2}^{n+2}{a}_{k-1}}\\ & \text{f}.{\displaystyle \sum _{k=1}^{n-5}{a}_k=\sum _{k=1}^n{a}_k-\sum _{k=(n-5)+1}^n{a}_k}\end{array}$

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
2. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.

Lanjutan 1 : Grafik Fungsi Trigonometri (Matematika Peminatan Kelas XI)

$\text{B. Grafik Fungsi Trigonometri}$

Dalam melukis grafik fungsi trigonometri nantinya yang ditampilkan adalah nilai-nilai untuk sudut istimewa saja. Dan selanjutnya besar sudutnya disajikan dalam derajat dan atau radian.

Selanjutnya perhatikanlah uraian berikut

$\text{1. Grafik Fungsi Sinus}$

$\begin{array}{|cccccccccc|}\hline x& 0& \frac{\pi }{6}& \frac{\pi }{4}& \frac{\pi }{3}& \frac{\pi }{2}& \frac{2\pi }{3}& \frac{3\pi }{4}& \frac{5\pi }{6}& \pi \\ f(x)& 0& \frac{1}{2}& \frac{1}{2}\sqrt{2}& \frac{1}{2}\sqrt{3}& 1& \frac{1}{2}\sqrt{3}& \frac{1}{2}\sqrt{2}& \frac{1}{2}& 0\\ x& \frac{7\pi }{6}& \frac{5\pi }{4}& \frac{4\pi }{3}& \frac{3\pi }{2}& \frac{5\pi }{3}& \frac{7\pi }{4}& \frac{11\pi }{6}& 2\pi & \\ f(x)& -\frac{1}{2}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}\sqrt{3}& -1& -\frac{1}{2}\sqrt{3}& -\frac{1}{2}\sqrt{2}& -\frac{1}{2}& 0& \\ \hline\end{array}$