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

$\begin{array}{l}\\ 101.& \text{Nilai dari}\sin {1020}^{\circ }=....\\ & \text{a}.-1\\ & \text{b}.{\displaystyle -\frac{1}{2}\sqrt{3}}\\ & \text{c}.{\displaystyle -\frac{1}{2}}\\ & \text{d}.{\displaystyle \frac{1}{2}}\\ & \text{e}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\sin {1020}^{\circ }& =\sin (3\times {360}^{\circ }-{60}^{\circ })\\ & =\sin (0^{\circ }-{60}^{\circ })\\ & =\sin (-{60}^{\circ })\\ & =-\sin {60}^{\circ }\\ & =-{\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\end{array}$.

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

$\begin{array}{l}\\ 103.& \text{Nilai dari}\\ & \sin {240}^{\circ }+\sin {225}^{\circ }+\cos {315}^{\circ }=....\\ & \begin{array}{l}\\ \text{a}.-\sqrt{3}& & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{b}.{\displaystyle -\frac{1}{2}\sqrt{3}}& & \text{e}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \text{c}.{\displaystyle -\frac{1}{2}}\end{array}\\ & \text{Jawab}:\text{b}\\ & \begin{array}{rl}& \sin {240}^{\circ }+\sin {225}^{\circ }+\cos {315}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })+\sin ({180}^{\circ }+{45}^{\circ })+\cos ({360}^{\circ }-{45}^{\circ })\\ & =-\sin {60}^{\circ }+[-\sin {45}^{\circ }]+\cos {45}^{\circ }\\ & =(-\frac{1}{2}\sqrt{3})+(-\frac{1}{2}\sqrt{2})+\frac{1}{2}\sqrt{2}\\ & =-\frac{1}{2}\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 104.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin {30}^{\circ }+\sin {150}^{\circ }+\cos {330}^{\circ }}{\tan {45}^{\circ }+\cos {210}^{\circ }}=....}\\ & \begin{array}{l}\\ \text{a}.{\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ \text{b}.{\displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}}\\ \text{c}.{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \text{d}.{\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}}\\ \text{e}.{\displaystyle \frac{1+2\sqrt{3}}{1-2\sqrt{3}}}\end{array}\\ & \text{Jawab}:\text{d}\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {30}^{\circ }+\sin {150}^{\circ }+\cos {330}^{\circ }}{\tan {45}^{\circ }+\cos {210}^{\circ }}}\\ & ={\displaystyle \frac{\sin {30}^{\circ }+\sin ({180}^{\circ }-{30}^{\circ })+\cos ({360}^{\circ }-{30}^{\circ })}{\tan {45}^{\circ }+\cos ({180}^{\circ }+{30}^{\circ })}}\\ & ={\displaystyle \frac{\sin {30}^{\circ }+\sin {30}^{\circ }+\cos {30}^{\circ }}{\tan {45}^{\circ }-\cos {30}^{\circ }}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sqrt{3}}}{1-{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{2}\sqrt{3}}}{1-{\displaystyle \frac{1}{2}\sqrt{3}}}}\\ & ={\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 105.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin {270}^{\circ }\times \cos {135}^{\circ }\times \tan {135}^{\circ }}{\sin {150}^{\circ }\times \cos {225}^{\circ }}=....}\\ & \begin{array}{l}\\ \text{a}.-2& & \text{d}.1\\ \text{b}.{\displaystyle -\frac{1}{2}}& \text{c}.{\displaystyle \frac{1}{2}}& \text{e}.2\end{array}\\ & \text{Jawab}:\text{e}\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {270}^{\circ }\times \cos {135}^{\circ }\times \tan {135}^{\circ }}{\sin {150}^{\circ }\times \cos {225}^{\circ }}}\\ & ={\displaystyle \frac{\sin {270}^{\circ }\times \cos ({180}^{\circ }-{45}^{\circ })\times \tan ({180}^{\circ }-{45}^{\circ })}{\sin ({180}^{\circ }-{30}^{\circ })\times \cos ({180}^{\circ }+{45}^{\circ })}}\\ & ={\displaystyle \frac{-1\times (-\cos {45}^{\circ })\times (-\tan {45}^{\circ })}{\sin {30}^{\circ }\times (-\cos {45}^{\circ })}}\\ & ={\displaystyle \frac{-1\times (-{\displaystyle \frac{1}{2}\sqrt{2}})\times -1}{{\displaystyle \frac{1}{2}\times (-\frac{1}{2}\sqrt{2})}}}\\ & ={\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{-\frac{1}{4}\sqrt{2}}}\\ & ={\displaystyle 2}\end{array}\end{array}$

$\begin{array}{l}\\ 106.& \text{Perhatikanlah gambar kurva berikut ini}\end{array}$.

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

$\begin{array}{l}\\ 107.& \text{Nilai minimum jika}\\ & f(x)={(2004\cos 2005x-2006)}^2+2007\\ & \text{adalah}=....\\ & \begin{array}{l}\\ \text{a}.& 2005\\ \text{b}.& 2007\\ \text{c}.& 2011\\ \text{d}.& 2013\\ \text{e}.& \text{tidak ada satupun jawaban dari a sampai d}\\ \\ & (\mathbf{\text{NUS Mathematics A Level}})\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& f(x)={(2004\cos 2005x-2006)}^2+2007\\ & \text{Supaya bernilai minimum, }\\ & \text{maka nilai}\cos 500x=1,\\ & \text{ingat nilai}-1\le \cos n\pi \le 1\\ & \text{maka},\\ & f_{min}={(2004.1-2006)}^2+2007\\ & =(-2{)}^2+2007\\ & =4+2007\\ & =2011\end{array}\end{array}$.

$\begin{array}{l}\\ 108.& \text{Penyelesaian persamaan}\\ & {\cos }^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \pi -{\cot }^{-1}\left({\displaystyle \frac{1}{2}}\right)\\ \text{b}.& \pi +{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)\\ \text{c}.& \pi -{\cot }^{-1}(-1)\\ \text{d}.& \pi +{\tan }^{-1}(-{\displaystyle \frac{1}{2}})\\ \text{e}.& \pi -{\tan }^{-1}\left({\displaystyle \frac{1}{4}}\right)\end{array}\\ & \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& \text{Perhatikan bahwa},\\ & {\cos }^{2}x-2\cos x=4\sin x-2\sin x\cos x\\ & \cos x(\cos x-2)=2\sin x(2-\cos x)\\ & \cos x(\cos x-2)=-2\sin x(\cos x-2)\\ & (\cos x+2\sin x)(\cos x-2)=0\\ & (\cos x+2\sin x)=0\text{atau}(\cos x-2)=0\\ & 2\sin x=-\cos x\mathbf{\text{(mm)}}\text{atau}\cos x=2\mathbf{\text{(tm)}}\\ & \text{maka}\\ & {\displaystyle \frac{\sin x}{\cos x}=-{\displaystyle \frac{1}{2}\iff \tan x=-{\displaystyle \frac{1}{2}}}}\\ & x={\tan }^{-1}(-{\displaystyle \frac{1}{2}})+k.\pi ,k\in \mathbb{Z}\end{array}\end{array}$.

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

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