Rumus Perkalian, Penjumlahan dan Pengurangan Sinus dan Cosinus

A. Rumus Perkalian Sinus dan Cosinus

Berikut rumus perkalian sinus dan cosinus:
1. $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )+\sin (\alpha -\beta ) \right]$
2. $\cos \alpha \sin \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )-\sin (\alpha -\beta ) \right]$
3. $\cos \alpha \cos \beta =\frac{1}{2}\left[ \cos (\alpha +\beta )+\cos (\alpha -\beta ) \right]$
4. $\sin \alpha \sin \beta =-\frac{1}{2}\left[ \cos (\alpha +\beta )-\cos (\alpha -\beta ) \right]$

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Contoh 1.
Tentukan nilai dari $6\cos 75^\circ \sin 15^\circ $.
Penyelesaian:
Ingat, $\cos \alpha \sin \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )-\sin (\alpha -\beta ) \right]$ maka:
$6\cos 75^\circ \sin 5^\circ $
= $6.\frac{1}{2}[\sin (75^\circ +15^\circ )-\sin (75^\circ -15^\circ )]$
= $3.[\sin (75^\circ +15^\circ )-\sin (75^\circ -15^\circ )]$
= $3.(\sin 90^\circ -\sin 60^\circ )$
= $3.\left( 1-\frac{1}{2}\sqrt{3} \right)$
= $3-\frac{3}{2}\sqrt{3}$

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Contoh 2.
Tentukan nilai dari $4\cos 37,5^\circ \cos 7,5^\circ $
Penyelesaian:
Ingat, $\cos \alpha \cos \beta =\frac{1}{2}\left[ \cos (\alpha +\beta )+\cos (\alpha -\beta ) \right]$ maka:
$4\cos 37,5^\circ \cos 7,5^\circ $
= $4.\frac{1}{2}$[$\cos (37,5^\circ +7,5^\circ)$+$\cos (37,5^\circ -7,5^\circ)$]
= $2.[\cos (37,5^\circ +7,5^\circ )+\cos (37,5^\circ -7,5^\circ )]$
= $2.(\cos 45^\circ +\cos 30^\circ )$
= $2.\left( \frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3} \right)$
= $\sqrt{2}+\sqrt{3}$

Contoh 3.
Hitunglah nilai dari $\sin 10^\circ \sin 50^\circ \sin 70^\circ $
Penyelesaian:
Ingat, $\sin \alpha \sin \beta =-\frac{1}{2}\left[ \cos (\alpha +\beta )-\cos (\alpha -\beta ) \right]$ dan $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )+\sin (\alpha -\beta ) \right]$ maka:
$\sin 10^\circ \sin 50^\circ \sin 70^\circ $
= $\sin 10^\circ (\sin 50^\circ \sin 70^\circ )$
= $\sin 10^\circ$.$\left[ -\frac{1}{2}\left[ \cos (50^\circ +70^\circ )-\cos (50^\circ -70^\circ ) \right] \right]$
= $\sin 10^\circ \left( -\frac{1}{2}.(\cos 120^\circ -\cos (-20^\circ )) \right)$
= $\sin 10^\circ \left( -\frac{1}{2}.(\cos 120^\circ -\cos 20^\circ ) \right)$
= $\sin 10^\circ \left( -\frac{1}{2}.\left( -\frac{1}{2}-\cos 20^\circ \right) \right)$
= $\sin 10^\circ \left( \frac{1}{4}+\frac{1}{2}\cos 20^\circ \right)$
= $\frac{1}{4}\sin 10^\circ +\frac{1}{2}\sin 10^\circ \cos 20^\circ $
= $\frac{1}{4}\sin 10^\circ$ + $\frac{1}{2}.\frac{1}{2}\left[ \sin (10^\circ +20^\circ )+\sin (10^\circ -20^\circ ) \right]$
= $\frac{1}{4}\sin 10^\circ +\frac{1}{4}(\sin 30^\circ +\sin (-10^\circ ))$
= $\frac{1}{4}\sin 10^\circ +\frac{1}{4}\left( \frac{1}{2}-\sin 10^\circ \right)$
= $\frac{1}{4}\sin 10^\circ +\frac{1}{8}-\frac{1}{4}\sin 10^\circ $
= $\frac{1}{8}$

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Contoh 4.
Tentukan nilai dari $-8\sin 142\frac{1}{2}^\circ \cos 97\frac{1}{2}^\circ $.
Penyelesaian:
Ingat, $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )+\sin (\alpha -\beta ) \right]$ maka:
$-8\sin 142\frac{1}{2}^\circ \cos 97\frac{1}{2}^\circ $
= $-8\frac{1}{2}$[$\sin (142\frac{1}{2}^\circ +97\frac{1}{2}^\circ)$+$\sin (142\frac{1}{2}^\circ -97\frac{1}{2}^\circ)$]
= $-4.\left( \sin 240^\circ +\sin 45^\circ \right)$
= $-4.\left( -\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2} \right)$
= $2\sqrt{3}-2\sqrt{2}$

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Contoh 5.
Tentukan himpunan penyelesaian dari persamaan $\sin \left( x+45^\circ \right)\cos x=\frac{1}{4}\sqrt{2}$ untuk $0^\circ \le x\le 360^\circ $.
Penyelesaian:
Ingat, $\sin \alpha \cos \beta =\frac{1}{2}\left[ \sin (\alpha +\beta )+\sin (\alpha -\beta ) \right]$ maka:
$\sin \left( x+45^\circ \right)\cos x = \frac{1}{4}\sqrt{2}$
$\frac{1}{2}\left[ \sin (x+45^\circ +x)+\sin (x+45^\circ -x) \right]$ = $\frac{1}{4}\sqrt{2}$
$\sin (2x+45^\circ )+\sin 45^\circ = \frac{1}{2}\sqrt{2}$
$\sin (2x+45^\circ )+\frac{1}{2}\sqrt{2} = \frac{1}{2}\sqrt{2}$
$\sin (2x+45^\circ ) = 0$
$\sin (2x+45^\circ ) = \sin 0^\circ$
Bentuk terakhir ini adalah persamaan trigonometri dasar $\sin f(x)=\sin g(x)$ dengan $f(x)=2x+45^\circ$ dan $g(x)=0^\circ$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x+45^\circ &= 0^\circ +k.360^\circ \\ 2x &= -45^\circ +k.360^\circ \\ x &= -22,5^\circ +k.180^\circ \end{align}$
$k=1\to x=157,5^\circ $
$k=2\to x=337,5^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x+45^\circ &= (180^\circ -0^\circ )+k.360^\circ \\ 2x &= 180^\circ -45^\circ +k.360^\circ \\ 2x &= 135^\circ +k.360^\circ \\ x &= 67,5^\circ +k.180^\circ \end{align}$
$k=0\to x=67,5^\circ $
$k=1\to x=247,5^\circ $
HP = $\{67,5^\circ ;\,157,5^\circ ;\,247,5^\circ ;\,337,5^\circ \}$

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B. Rumus Penjumlahan dan Pengurangan Sinus dan Cosinus

Berikut ini rumus penjumlahan/pengurangan fungsi trigonometri:
1. $\sin \alpha +\sin \beta$ = $2\sin \frac{1}{2}(\alpha +\beta )\cos \frac{1}{2}(\alpha -\beta )$
2. $\sin \alpha -\sin \beta$ = $2\cos \frac{1}{2}(\alpha +\beta )\sin \frac{1}{2}(\alpha -\beta )$
3. $\cos \alpha +\cos \beta$ = $2\cos \frac{1}{2}(\alpha +\beta )\cos \frac{1}{2}(\alpha -\beta )$
4. $\cos \alpha -\cos \beta$ = $-2\sin \frac{1}{2}(\alpha +\beta )\sin \frac{1}{2}(\alpha -\beta )$

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Contoh 1.
Tentukan nilai dari $\sin 75^\circ -\sin 165^\circ $.
Penyelesaian:
Ingat:
$\sin \alpha -\sin \beta =2\cos \frac{1}{2}(\alpha +\beta )\sin \frac{1}{2}(\alpha -\beta )$ maka:
$\sin 75^\circ -\sin 165^\circ $
$=2\cos \frac{1}{2}(75^\circ +165^\circ )\sin \frac{1}{2}(75^\circ -165^\circ )$
$=2\cos 120^\circ \sin (-45^\circ )$
$=2\times \left( -\frac{1}{2} \right).\left( -\frac{1}{2}\sqrt{2} \right)$
$=\frac{1}{2}\sqrt{2}$

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Contoh 2.
Tentukan nilai dari $\cos 25^\circ +\cos 95^\circ +\cos 145^\circ $.
Penyelesaian:
Ingat:
$\cos \alpha +\cos \beta =2\cos \frac{1}{2}(\alpha +\beta )\cos \frac{1}{2}(\alpha -\beta )$ maka:
$\cos 25^\circ +\cos 95^\circ +\cos 145^\circ $
= $\cos 25^\circ +(\cos 95^\circ +\cos 145^\circ )$
= $\cos 25^\circ$ + $2\cos \frac{1}{2}(95^\circ +145^\circ )\cos \frac{1}{2}(95^\circ -145^\circ )$
= $\cos 25^\circ +2\cos 120^\circ \cos (-25^\circ )$
= $\cos 25^\circ +2\times \left( -\frac{1}{2} \right)\cos 25^\circ $
= $\cos 25^\circ -\cos 25^\circ $
= 0

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Contoh 3.
Tentukan nilai dari $\frac{\sin 81^\circ +\sin 21^\circ }{\sin 69^\circ -\sin 171^\circ }$.
Penyelesaian:
$\frac{\sin 81^\circ +\sin 21^\circ }{\sin 69^\circ -\sin 171^\circ }$
= $\frac{2\sin \frac{1}{2}(81^\circ +21^\circ )\cos \frac{1}{2}(81^\circ -21^\circ )}{2\cos \frac{1}{2}(69^\circ +171^\circ )\sin \frac{1}{2}(69^\circ -171^\circ )}$
= $\frac{2\sin 51^\circ \cos 30^\circ }{2\cos 120^\circ \sin (-51^\circ )}$
= $\frac{\sin 51^\circ \cos 30^\circ }{-\sin 51^\circ \cos 120^\circ }$
= $-\frac{\cos 30^\circ }{\cos 120^\circ }$
= $-\frac{\frac{1}{2}\sqrt{3}}{-\frac{1}{2}}$
= $\sqrt{3}$

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Contoh 4.
Tentukan nilai dari $\frac{\cos 140^\circ -\cos 100^\circ }{\sin 140^\circ -\sin 100^\circ }$.
Penyelesaian:
$\frac{\cos 140^\circ -\cos 100^\circ }{\sin 140^\circ -\sin 100^\circ }$
$=\frac{-2\sin \frac{1}{2}(140^\circ +100^\circ )\sin \frac{1}{2}(140^\circ -100^\circ )}{2\cos \frac{1}{2}(140^\circ +100^\circ )\sin \frac{1}{2}(140^\circ -100^\circ )}$
= $-\frac{\sin 120^\circ \sin 20^\circ }{\cos 120^\circ \sin 20^\circ }$
= $-\frac{\sin 120^\circ }{\cos 120^\circ }$
= $-\frac{\frac{1}{2}\sqrt{3}}{-\frac{1}{2}}$
= $\sqrt{3}$

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Contoh 5.
Tentukan himpunan penyelesaian dari persamaan $\cos (2x+150^\circ )+\cos (2x-90^\circ )=\frac{1}{2}\sqrt{3}$ untuk $0^\circ \le x\le 360^\circ $.
Penyelesaian:
Ingat:
$\cos \alpha +\cos \beta =2\cos \frac{1}{2}(\alpha +\beta )\cos \frac{1}{2}(\alpha -\beta )$ maka:
$\begin{align}\cos (2x+150^\circ )+\cos (2x-90^\circ ) &= \frac{1}{2}\sqrt{3} \\ 2\cos \frac{1}{2}(4x+60^\circ )\cos (\frac{1}{2}.240^\circ ) &= \frac{1}{2}\sqrt{3} \\ 2\cos (2x+30^\circ )\cos 120^\circ &= \frac{1}{2}\sqrt{3} \\ 2\cos (2x+30^\circ )\times \left( -\frac{1}{2} \right) &= \frac{1}{2}\sqrt{3} \\ -\cos (2x+30^\circ ) &= \frac{1}{2}\sqrt{3} \\ \cos (2x+30^\circ ) &= -\frac{1}{2}\sqrt{3} \\ \cos (2x+30^\circ ) &= \cos 150^\circ \end{align}$
Bentuk terakhir ini adalah persamaan trigonometri dasar $\cos f(x)=\cos g(x)$ dengan $f(x)=2x+30^\circ$ dan $g(x)=150^\circ$ maka:
1) $f(x)=g(x)+k.360^\circ $
$\begin{align}2x+30^\circ &= 150^\circ +k.360^\circ \\ 2x &= 150^\circ -30^\circ +k.360^\circ \\ 2x &= 120^\circ +k.360^\circ \\ x &= 60^\circ +k.180^\circ \end{align}$
$k=0\to x=60^\circ $
$k=1\to x=240^\circ $
2) $f(x)=(180^\circ -g(x))+k.360^\circ $
$\begin{align}2x+30^\circ &= (180^\circ -150^\circ )+k.360^\circ \\ 2x+30^\circ &= 30^\circ +k.360^\circ \\ 2x &= 30^\circ -30^\circ +k.360^\circ \\ 2x &= k.360^\circ \\ x &= k.180^\circ \end{align}$
$k=0\to x=0^\circ $
$k=1\to x=180^\circ $
$k=2\to x=360^\circ $
HP = $\{0^\circ ,60^\circ ,180^\circ ,240^\circ ,360^\circ \}$

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C. Soal Latihan

1. Tentukanlah nilai dari $\cos 97\frac{1}{2}^\circ \sin 67\frac{1}{2}^\circ $.
2. Tentukanlah nilai dari $-12\tan 20^\circ .\tan 40^\circ .\tan 80^\circ $.
3. Tentukan nilai dari $\sin 50^\circ -\sin 70^\circ +\sin 10^\circ $.
4. Tentukan hasil dari $\frac{\sin 27^\circ +\sin 63^\circ }{\cos 138^\circ +\cos 102^\circ }$.
5. Tentukan himpunan penyelesaian dari persamaan $\sin (3x-150^\circ )-\sin (3x-30^\circ )=-\sqrt{3}$ untuk $0^\circ \le x\le 360^\circ $.

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