Rumus Trigonometri Sudut Ganda dan Sudut Pertengahan

A. Rumus Trigonometri Sudut Ganda

1. $\sin 2\alpha =2\sin \alpha \cos \alpha $
2. $\cos 2\alpha ={{\cos }^2}\alpha -{{\sin }^2}\alpha $
3. $\cos 2\alpha =1-2{{\sin }^2}\alpha $
4. $\cos 2\alpha =2{{\cos }^2}\alpha -1$
5. $\tan 2\alpha =\frac{2\tan \alpha }{1-{{\tan }^2}\alpha }$ dengan $\alpha \ne (2k-1)\frac{\pi }{2}$, $k\in B$.

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Contoh 1.
Diketahui $\tan \alpha =p$ dan $\alpha $ sudut lancip, tentukan $\sin 2\alpha $ dan $\cos 2\alpha $.
Penyelesaian:
$\tan \alpha =p$
$\tan \alpha =\frac{p}{1}=\frac{de}{sa}$
$mi=\sqrt{de^2+sa^2}\Leftrightarrow mi=\sqrt{p^2+1}$
$\sin \alpha =\frac{de}{mi}\Leftrightarrow \sin \alpha =\frac{p}{\sqrt{p^2+1}}$
$\cos \alpha =\frac{sa}{mi}\Leftrightarrow \cos \alpha =\frac{1}{\sqrt{p^2+1}}$
$\begin{align}\sin 2\alpha &= 2\sin \alpha \cos \alpha \\ &= 2.\frac{p}{\sqrt{p^2+1}}.\frac{1}{\sqrt{p^2+1}} \\ \sin 2\alpha &= \frac{2p}{p^2+1} \end{align}$
$\begin{align}\cos 2\alpha &= {\cos }^2\alpha -{\sin }^2\alpha \\ &= {{\left( \frac{1}{\sqrt{p^2+1}} \right)}^2}-{{\left( \frac{p}{\sqrt{p^2+1}} \right)}^2} \\ &= \frac{1}{p^2+1}-\frac{p^2}{p^2+1} \\ \cos 2\alpha &= \frac{1-p^2}{p^2+1} \end{align}$

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Contoh 2.
Diketahui $\sin p=\frac{2}{\sqrt{5}}$ dan $0^\circ < p < 90^\circ $, tentukan nilai dari $\tan 2p$.
Penyelesaian:
$\sin p=\frac{2}{\sqrt{5}}=\frac{de}{mi}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{(\sqrt{5})^2-2^2} \\ sa &= 1 \end{align}$
$\tan p=\frac{de}{sa}\Leftrightarrow \tan p=\frac{2}{1}=2$
$\begin{align}\tan 2p &= \frac{2\tan p}{1-{{\tan }^2}p} \\ &= \frac{2.2}{1-2^2} \\ \tan 2p &= -\frac{4}{3} \end{align}$

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Contoh 3.
Ditentukan ${{\sin }^2}A=\frac{3}{5}$ untuk $\frac{\pi }{2} < A < \pi $, tentukan nilai $\tan 2A$.
Penyelesaian:
$\frac{\pi }{2} < A < \pi $ maka $\sin A > 0$ dan $\tan A < 0$
${{\sin }^2}A=\frac{3}{5}\Leftrightarrow \sin A=\frac{\sqrt{3}}{\sqrt{5}}=\frac{de}{mi}$
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{(\sqrt{5})^2-(\sqrt{3})^2} \\ sa &= \sqrt{2} \end{align}$
$\tan A=-\frac{de}{sa}\Leftrightarrow \tan A=-\frac{\sqrt{3}}{\sqrt{2}}$
$\begin{align}\tan 2A &= \frac{2\tan A}{1-{\tan }^2A} \\ &= \frac{2.\left( -\frac{\sqrt{3}}{\sqrt{2}} \right)}{1-{{\left( -\frac{\sqrt{3}}{\sqrt{2}} \right)}^2}} \\ &= \frac{-\frac{2\sqrt{3}}{\sqrt{2}}}{1-\frac{3}{2}} \\ &= \frac{-\frac{2\sqrt{3}}{\sqrt{2}}}{-\frac{1}{2}} \\ &= -\frac{2\sqrt{3}}{\sqrt{2}}\times \left(-\frac{2}{1}\right) \\ &= \frac{4\sqrt{3}}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} \\ \tan 2A &= 2\sqrt{6} \end{align}$

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Contoh 4.
Ditentukan $\tan \frac{1}{2}A=t$, tentukan $\sin A$.
Penyelesaian:
$\tan \frac{1}{2}A=t$
$\tan \frac{1}{2}A=\frac{t}{1}=\frac{de}{sa}$
$mi=\sqrt{de^2+sa^2}\Leftrightarrow m=\sqrt{t^2+1}$
$\sin \frac{1}{2}A=\frac{de}{mi}=\frac{t}{\sqrt{t^2+1}}$
$\cos \frac{1}{2}A=\frac{sa}{mi}=\frac{1}{\sqrt{t^2+1}}$
$\begin{align}\sin A &= \sin 2\left( \frac{1}{2}A \right) \\ &= 2\sin \frac{1}{2}A.\cos \frac{1}{2}A \\ &= 2.\frac{t}{\sqrt{t^2+1}}.\frac{1}{\sqrt{t^2+1}} \\ \sin A &= \frac{2t}{t^2+1} \end{align}$

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Contoh 5.
Tentukan himpunan penyelesaian dari persamaan $\cos 4x+3\sin 2x=-1$ untuk $0^\circ < x < 360^\circ $.
Penyelesaian:
$\begin{align}\cos 4x+3\sin 2x &= -1 \\ \cos (2.2x)+3\sin 2x &= -1 \\ 1-2{{\sin }^2}2x+3\sin 2x &= -1 \\ -2{{\sin }^2}2x+3\sin 2x+2 &= 0 \\ 2{{\sin }^2}2x-3\sin 2x-2 &= 0 \\ (\sin 2x-2)(2\sin 2x+1) &= 0 \end{align}$
$\sin 2x-2=0\Leftrightarrow \sin 2x=2$
(ditolak, karena nilai maksimum $\sin 2x$ adalah 1)
$\begin{align}2\sin 2x+1 &= 0 \\ \sin 2x &= -\frac{1}{2} \\ \sin 2x &= \sin 210^\circ \end{align}$
Diperoleh persamaan trigonometri dasar yaitu $\sin f(x)=\sin g(x)$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x &= 210^\circ +k.360^\circ \\ x &= 105^\circ +k.180^\circ \end{align}$
$k=0\to x=105^\circ $
$k=1\to x=285^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x &= (180^\circ -210^\circ )+k.360^\circ \\ 2x &= -30^\circ +k.360^\circ \\ x &= -15^\circ +k.180^\circ \end{align}$
$k=1\to x=165^\circ $
$k=2\to x=345^\circ $
HP = $\{105^\circ ,165^\circ ,285^\circ ,345^\circ \}$

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B. Rumus Trigonometri Sudut Pertengahan

1. $\sin \frac{1}{2}\alpha =\pm \sqrt{\frac{1-\cos \alpha }{2}}$, tanda positif (+) untuk $\frac{1}{2}\alpha $ di kuadran I dan II dan tanda negatif (–) untuk $\frac{1}{2}\alpha $ di kuadran III dan IV.
2. $\cos \frac{1}{2}\alpha =\pm \sqrt{\frac{1+\cos \alpha }{2}}$, tanda positif (+) untuk $\frac{1}{2}\alpha $ di kuadran I dan IV dan tanda negatif (–) untuk $\frac{1}{2}\alpha $ di kuadran II dan III.
3. $\tan \frac{1}{2}\alpha =\pm \sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}$, tanda positif (+) untuk $\frac{1}{2}\alpha $ di kuadran I dan IIL dan tanda negatif (–) untuk $\frac{1}{2}\alpha $ di kuadran II dan IV.
4. $\tan \frac{1}{2}\alpha =\frac{\sin \alpha }{1+\cos \alpha }=\frac{1-\cos \alpha }{\sin \alpha }$

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Contoh 1.
Tanpa menggunakan kalkulator hitunglah nilai dari $\sin 22\frac{1}{2}^\circ $, $\cos 157\frac{1}{2}^\circ $ dan $\tan 105^\circ $.
Penyelesaian:
$\begin{align}\sin 22\frac{1}{2}^\circ &= \sin \frac{1}{2}\times 45^\circ \\ &= \sqrt{\frac{1-\cos 45^\circ }{2}} \\ &= \sqrt{\frac{1-\frac{1}{2}\sqrt{2}}{2}} \\ &= \sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}} \\ &= \sqrt{\frac{2-\sqrt{2}}{4}} \\ \sin 22\frac{1}{2}^\circ &= \frac{\sqrt{2-\sqrt{2}}}{2} \end{align}$

$\begin{align}\cos 157\frac{1}{2}^\circ &= \cos \frac{1}{2}\times 315^\circ \\ &= -\sqrt{\frac{1+\cos 315^\circ }{2}} \\ &= -\sqrt{\frac{1+\cos (360^\circ -45^\circ )}{2}} \\ &= -\sqrt{\frac{1+\cos 45^\circ }{2}} \\ &= -\sqrt{\frac{1+\frac{1}{2}\sqrt{2}}{2}} \\ &= -\sqrt{\frac{\frac{2+\sqrt{2}}{2}}{2}} \\ &= -\sqrt{\frac{2+\sqrt{2}}{4}} \\ \cos 157\frac{1}{2}^\circ &= -\frac{\sqrt{2+\sqrt{2}}}{2} \end{align}$

$\begin{align}\tan 105^\circ &= \tan \frac{1}{2}\times 210^\circ \\ &= \frac{\sin 210^\circ }{1+\cos 210^\circ } \\ &= \frac{\sin (180^\circ +30^\circ )}{1+\cos (180^\circ +30^\circ )} \\ &= \frac{-\sin 30^\circ }{1-\cos 30^\circ } \\ &= \frac{-\frac{1}{2}}{1-\frac{1}{2}\sqrt{3}} \\ &= \frac{-\frac{1}{2}}{\frac{2-\sqrt{3}}{2}} \\ &= \frac{-1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\ &= \frac{-2-\sqrt{3}}{4-3} \\ \tan 105^\circ &= -2-\sqrt{3} \end{align}$

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Contoh 2.
Diketahui $\cos \alpha =\frac{3}{5}$ dengan $\alpha $ sudut lancip. Hitunglah nilai $\sin \frac{1}{2}\alpha $, $\cos \frac{1}{2}\alpha $ dan $\tan \frac{1}{2}\alpha $.
Penyelesaian:
$\cos \alpha =\frac{3}{5}$
$\begin{align}\sin \frac{1}{2}\alpha &= \sqrt{\frac{1-\cos \alpha }{2}} \\ &= \sqrt{\frac{1-\frac{3}{5}}{2}} \\ &= \sqrt{\frac{1}{5}} \\ &= \frac{1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ \sin \frac{1}{2}\alpha &= \frac{1}{5}\sqrt{5} \end{align}$

$\begin{align}\cos \frac{1}{2}\alpha &= \sqrt{\frac{1+\cos \alpha }{2}} \\ &= \sqrt{\frac{1+\frac{3}{5}}{2}} \\ &= \sqrt{\frac{4}{5}} \\ &= \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ \cos \frac{1}{2}\alpha &= \frac{2}{5}\sqrt{5} \end{align}$

$\begin{align}\tan \frac{1}{2}\alpha &= \frac{\sin \frac{1}{2}\alpha }{\cos \frac{1}{2}\alpha } \\ &= \frac{\frac{1}{5}\sqrt{5}}{\frac{2}{5}\sqrt{5}} \\ \tan \frac{1}{2}\alpha &= \frac{1}{2} \end{align}$

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Contoh 3.
Diketahui $\sin 3x=\frac{5}{13}$, hitunglah nilai dari $\sin \frac{3}{2}x$.
Penyelesaian:
$\sin 3x=\frac{5}{13}=\frac{de}{mi}$ maka
$\begin{align}sa &= \sqrt{mi^2-de^2} \\ &= \sqrt{13^2-5^2} \\ sa &= 12 \end{align}$
$\cos 3x=\frac{sa}{mi}\Leftrightarrow \cos 3x=\frac{12}{13}$
$\begin{align}\sin \frac{3}{2}x &= \sin \left( \frac{1}{2}.3x \right) \\ &= \sqrt{\frac{1-\cos 3x}{2}} \\ &= \sqrt{\frac{1-\frac{12}{13}}{2}} \\ &= \sqrt{\frac{1}{26}} \\ &= \frac{1}{\sqrt{26}}\times \frac{\sqrt{26}}{\sqrt{26}} \\ \sin \frac{3}{2}x &= \frac{1}{26}\sqrt{26} \end{align}$

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C. Soal Latihan

1. Hitunglah nilai $2\sin \frac{\pi }{12}\cos \frac{\pi }{12}$.
2. Diketahui $\tan 2x=p$, tentukan $\tan 4x$.
3. Tentukan himpunan penyelesaian dari persamaan $\cos 2x-2\cos x=-1$ untuk $0^\circ < x < 360^\circ $.
4. Diketahui $\tan x=\frac{1}{4}$ untuk $\pi < x < \frac{3}{2}\pi $, tentukan nilai $\sin \frac{1}{2}x$.
5. Diketahui A sudut lancip dengan cos 2A = $\tfrac{1}{3}$, tentukan nilai tan A.

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