Rumus Trigonometri Jumlah dan Selisih Dua Sudut

A. Rumus Trionometri Jumlah dan Selisih Dua Sudut

1. $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta $
2. $\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta $
3. $\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta $
4. $\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta $
5. $\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
6. $\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$

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Contoh 1.
Tanpa menggunakan kalkulator hitunglah nilai dari $\sin 75^\circ $ dan $\sin 345^\circ $.
Penyelesaian:
$\begin{align}\sin 75^\circ &= \sin (30^\circ +45^\circ ) \\ &= \sin 30^\circ \cos 45^\circ +\cos 30^\circ \sin 45^\circ \\ &= \frac{1}{2}.\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2} \\ &= \frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{6} \\ \sin 75^\circ &= \frac{\sqrt{2}+\sqrt{6}}{4} \end{align}$

$\begin{align}\sin 345^\circ &= \sin (360^\circ -15^\circ ) \\ &= -\sin 15^\circ \\ &= \sin (-15^\circ ) \\ &= \sin (30^\circ -45^\circ ) \\ &= \sin 30^\circ \cos 45^\circ -\cos 30^\circ \sin 45^\circ \\ &= \frac{1}{2}.\frac{1}{2}\sqrt{2}-\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2} \\ &= \frac{1}{4}\sqrt{2}-\frac{1}{4}\sqrt{6} \\ \sin 345^\circ &= \frac{\sqrt{2}-\sqrt{6}}{4} \end{align}$

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Contoh 2.
Tanpa menggunakan kalkulator hitunglah nilai dari $\cos 15^\circ $ dan $\cos 285^\circ $.
Penyelesaian:
$\begin{align}\cos 15^\circ &= \cos (45-30) \\ &= \cos 45^\circ \cos 30^\circ +\sin 45^\circ \sin 30^\circ \\ &= \frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2}.\frac{1}{2} \\ &= \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2} \\ \cos 15^\circ &= \frac{\sqrt{6}+\sqrt{2}}{4} \end{align}$

$\begin{align}\cos 285^\circ &= \cos (360^\circ -75^\circ ) \\ &= \cos 75^\circ \\ &= \cos (45^\circ +30^\circ ) \\ &= \cos 45^\circ \cos 30^\circ -\sin 45^\circ \sin 30^\circ \\ &= \frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}.\frac{1}{2} \\ &= \frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{2} \\ \cos 285^\circ &= \frac{\sqrt{6}-\sqrt{2}}{4} \end{align}$

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Contoh 3.
Tanpa menggunakan kalkulator hitunglah nilai dari $\tan 105^\circ $ dan $\tan 195^\circ $.
Penyelesaian:
$\begin{align}\tan 105^\circ &= \tan (60^\circ +45^\circ ) \\ &= \frac{\tan 60^\circ +\tan 45^\circ }{1-\tan 60^\circ \tan 45^\circ } \\ &= \frac{\sqrt{3}+1}{1-\sqrt{3}.1} \\ &= \frac{\sqrt{3}+1}{1-\sqrt{3}}\times \frac{1+\sqrt{3}}{1+\sqrt{3}} \\ &= \frac{\sqrt{3}+3+1+\sqrt{3}}{1-3} \\ &= \frac{4+2\sqrt{3}}{-2} \\ \tan 105^\circ &= -2-\sqrt{3} \end{align}$

$\begin{align}\tan 195^\circ &= \tan (180^\circ +15^\circ ) \\ &= \tan 15^\circ \\ &= \tan (45^\circ -30^\circ ) \\ &= \frac{\tan 45^\circ -\tan 30^\circ }{1+\tan 45^\circ \tan 30^\circ } \\ &= \frac{1-\frac{1}{\sqrt{3}}}{1+1.\frac{1}{\sqrt{3}}} \\ &= \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} \\ &= \frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1} \\ &= \frac{3-\sqrt{3}-\sqrt{3}+1}{3-1} \\ &= \frac{4-2\sqrt{3}}{2} \\ \tan 195^\circ &= 2-\sqrt{3} \end{align}$

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Contoh 4.
Pada segitiga ABC lancip, diketahui $\cos A=\frac{4}{5}$ dan $\sin B=\frac{12}{13}$. Tentukan nilai $\sin C$.
Penyelesaian:
$\cos A=\frac{4}{5}=\frac{sa}{mi}$ maka:
$\begin{align}\sin A &= \frac{de}{mi} \\ &= \frac{\sqrt{mi^2-sa^2}}{mi} \\ &= \frac{\sqrt{5^2-4^2}}{5} \\ \sin A &= \frac{3}{5} \end{align}$
$\sin B=\frac{12}{13}=\frac{de}{mi}$ maka:
$\begin{align}\cos B &= \frac{sa}{mi} \\ &= \frac{\sqrt{13^2-12^2}}{13} \\ \cos B &= \frac{5}{13} \end{align}$
segitiga ABC maka:
$\begin{align}A+B+C &= 180^\circ \\ C &= 180^\circ-(A+B) \\ \sin C &= \sin (180^\circ-(A+B)) \\ &= \sin (A+B) \\ &= \sin A\cos B+\cos A\sin B \\ &= \frac{3}{5}.\frac{5}{13}+\frac{4}{5}.\frac{12}{13} \\ &= \frac{15}{65}+\frac{48}{65} \\ \sin C &= \frac{63}{65} \end{align}$

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Contoh 5.
Diketahui $(\alpha +\beta )=\frac{\pi }{3}$ dan $\sin \alpha \sin \beta =\frac{1}{4}$. Tentukan nilai dari $\cos (\alpha -\beta )$.
Penyelesaian:
$\begin{align}\cos (\alpha +\beta ) &= \cos \frac{\pi }{3} \\ \cos \alpha \cos \beta -\sin \alpha \sin \beta &= \frac{1}{2} \\ \cos \alpha \cos \beta -\frac{1}{4} &= \frac{1}{2} \\ \cos \alpha \cos \beta &= \frac{1}{2}+\frac{1}{4} \\ \cos \alpha \cos \beta &= \frac{3}{4} \end{align}$
$\begin{align}\cos (\alpha -\beta ) &= \cos \alpha \cos \beta +\sin \alpha \sin \beta \\ &= \frac{3}{4}+\frac{1}{4} \\ \cos (\alpha -\beta ) &= 1 \end{align}$

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Contoh 6.
Diketahui $\cos \alpha =\frac{3}{5}$, $\alpha $ adalah sudut lancip dan $\sin \beta =\frac{7}{25}$, $\beta $ adalah sudut tumpul. Tentukan nilai $\tan (\alpha +\beta )$.
Penyelesaian:
$\cos \alpha =\frac{3}{5}=\frac{sa}{mi}$
$\alpha $ adalah sudut lancip maka $\tan \alpha $ bernilai positif, maka:
$\begin{align}\tan \alpha &= \frac{de}{sa} \\ &= \frac{\sqrt{mi^2-sa^2}}{sa} \\ &= \frac{\sqrt{5^2-3^2}}{3} \\ \tan \alpha &= \frac{4}{3} \end{align}$

$\sin \beta =\frac{7}{25}=\frac{de}{mi}$
$\beta $ adalah sudut tumpul maka $\tan \beta $ bernilai negatif, maka:
$\begin{align}\tan \beta &= -\frac{de}{sa} \\ &= -\frac{de}{\sqrt{mi^2-de^2}} \\ &= -\frac{7}{\sqrt{25^2-7^2}} \\ &= -\frac{7}{\sqrt{25^2-7^2}} \\ \tan \beta &= -\frac{7}{24} \end{align}$
$\begin{align}\tan (\alpha +\beta ) &= \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } \\ &= \frac{\frac{4}{3}+\left( -\frac{7}{24} \right)}{1-\frac{4}{3}.\left( -\frac{7}{24} \right)} \\ &= \frac{\frac{32}{24}-\frac{7}{24}}{1+\frac{7}{18}} \\ &= \frac{\frac{25}{24}}{\frac{25}{18}} \\ &= \frac{25}{24}\times \frac{18}{25} \\ \tan (\alpha +\beta ) &=\frac{3}{4} \end{align}$

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Contoh 7.
Diketahui $\tan \alpha -\tan \beta =\frac{1}{3}$ dan $\cos \alpha \cos \beta =\frac{48}{65}$, $\alpha $ dan $\beta $ adalah sudut lancip. Tentukan nilai $\sin (\alpha -\beta )$.
Penyelesaian:
$\begin{align}\tan \alpha -\tan \beta &=\frac{1}{3} \\ \frac{\sin \alpha }{\cos \alpha }-\frac{\sin \beta }{\cos \beta } &= \frac{1}{3} \\ \frac{\sin \alpha \cos \beta -\cos \alpha \sin \beta }{\cos \alpha \cos \beta } &= \frac{1}{3} \\ \frac{\sin (\alpha -\beta )}{\frac{48}{65}} &= \frac{1}{3} \\ \sin (\alpha -\beta ) &= \frac{1}{3}\times \frac{48}{65} \\ \sin (\alpha -\beta ) &= \frac{16}{65} \end{align}$

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B. Soal Latihan

1. Diketahui $x$ dan $y$ adalah sudut lancip dan $x-y=30^\circ $. Jika $\cos x\sin y=\frac{1}{6}$ tentukan nilai dari $\sin x\cos y$.
2. Diketahui $\cos (A-B)=\frac{3}{5}$ dan $\cos A\cos B=\frac{7}{25}$. Tentukan nilai $\tan A\tan B$.
3. Diketahui $\cos A=\frac{2}{3}$, $\cos B=\frac{2}{5}$, A dan B adalah sudut lancip. Tentukan nilai dari $\cos (A+B)$.
4. Bila $\sin \alpha =\frac{5}{13}$ dan $\cos \beta =\frac{4}{5}$ dengan $\alpha $ dan $\beta $ adalah sudut lancip. Tentukan nilai dari $\tan (\alpha +\beta )$.
5. Pada suatu segitiga ABC yang siku-siku di C, diketahui bahwa $\sin A\sin B=\frac{2}{5}$ dan $\sin (A-B)=5a$. Tentukan nilai $a$.

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