C. Rumus Perkalian Sinus dan Cosinus
Untuk sudut $\alpha \: \: \textrm{dan}\: \: \beta$ berlaku rumus-rumus
$\begin{aligned}&\begin{cases} 2\sin \alpha \cos \gamma &=\sin \left ( \alpha +\gamma \right )+\sin \left ( \alpha -\gamma \right )\\ 2\cos \alpha \sin \gamma &=\sin \left ( \alpha +\gamma \right )-\sin \left ( \alpha -\gamma \right ) \end{cases}\\ &\begin{cases} 2\cos \alpha \cos \gamma &=\cos \left ( \alpha +\gamma \right )+\cos \left ( \alpha -\gamma \right )\\ - 2\sin \alpha \sin \gamma &=\cos \left ( \alpha +\gamma \right )-\cos \left ( \alpha -\gamma \right ) \end{cases} \end{aligned}$.
Sebagai buktinya akan ditunjukkan pada tulisan berikut
Bukti untuk nomor pertama
$\begin{aligned}\sin &\color{blue}\left ( \alpha +\beta \right )+\sin \left ( \alpha -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &\quad +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ &=\color{red}2\sin \alpha \cos \beta \end{aligned}$.
Selanjutnya untuk bukti baris kedua yaitu:
$\begin{aligned}\sin &\color{blue}\left ( \alpha +\beta \right )-\sin \left ( \alpha -\beta \right )\\ &=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ &\quad -\left (\sin \alpha \cos \beta -\cos \alpha \sin \beta \right ) \\ &=\color{red}2\cos \alpha \sin \beta \end{aligned}$.
Dan bukti untuk rumus ketiga
$\begin{aligned}\cos &\color{blue}\left ( \alpha +\beta \right )+\cos \left ( \alpha -\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &\quad +\cos \alpha \cos \beta+\sin \alpha \sin \beta \\ &=\color{red}2\cos \alpha \cos \beta \end{aligned}$.
Adapun bukti untuk rumus yang tertakhir adalah:
$\begin{aligned}\cos &\color{blue}\left ( \alpha +\beta \right )-\cos \left ( \alpha -\beta \right )\\ &=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ &\quad -\left (\cos \alpha \cos \beta +\sin \alpha \sin \beta \right ) \\ &=\color{red}-2\sin \alpha \sin \beta \end{aligned}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Nyatakanlah ke dalam bentuk jumlah}\\ &\textrm{atau selisih dari bentuk berikut}\\ &\textrm{a}.\quad 2\cos 80^{\circ}\sin 50^{\circ}\\ &\textrm{b}.\quad 2\sin 80^{\circ}\cos 50^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos 80^{\circ}\sin 50^{\circ}\\ &=\sin \left ( 80^{\circ}+50^{\circ} \right )-\sin \left ( 80^{\circ}-50^{\circ} \right )\\ &=\sin 130^{\circ}-\sin 30^{\circ}\\ &=\sin 130^{\circ}-\displaystyle \frac{1}{2}\\ \textrm{b}.\quad&2\sin 80^{\circ}\cos 50^{\circ}\\ &=\sin \left ( 80^{\circ}+50^{\circ} \right )+\sin \left ( 80^{\circ}-50^{\circ} \right )\\ &=\sin 130^{\circ}+\sin 30^{\circ}\\ &=\sin 130^{\circ}+\displaystyle \frac{1}{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Nyatakanlah ke dalam bentuk jumlah}\\ &\textrm{atau selisih dari bentuk berikut}\\ &\textrm{a}.\quad 2\cos 75^{\circ}\cos 15^{\circ}\\ &\textrm{b}.\quad \sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos 75^{\circ}\cos 15^{\circ}\\ &=\cos \left ( 75^{\circ}+15^{\circ} \right )+\cos \left ( 75^{\circ}-15^{\circ} \right )\\ &=\cos 90^{\circ}+\cos 60^{\circ}\\ &=0+\displaystyle \frac{1}{2}=\color{red}\displaystyle \frac{1}{2}\\ \textrm{b}.\quad&\sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right )\\ &=-\displaystyle \frac{1}{2}\left ( 2\sin \left ( \displaystyle \frac{3\pi }{8} \right )\sin \left ( \displaystyle \frac{\pi }{8} \right ) \right )\\ &=-\displaystyle \frac{1}{2}\left (\cos \left ( \displaystyle \frac{3\pi }{8}+\frac{\pi }{8} \right )-\cos \left ( \displaystyle \frac{3\pi }{8}-\frac{\pi }{8} \right ) \right )\\ &=-\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{4\pi }{8}-\cos \displaystyle \frac{2\pi }{8} \right )\\ &=-\displaystyle \frac{1}{2}\left ( \cos \displaystyle \frac{1}{2}\pi -\cos \displaystyle \frac{1}{4}\pi \right )\\ &=-\displaystyle \frac{1}{2}\left ( 0-\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=-\displaystyle \frac{1}{2}\left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{4}\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari bentuk berikut}\\\\ &\textrm{a}.\quad 2\sin 37\displaystyle \frac{1}{2}^{\circ}\cos 7\displaystyle \frac{1}{2}^{\circ}\\\\ &\textrm{b}.\quad 2\cos 82\displaystyle \frac{1}{2}^{\circ}\sin 37\displaystyle \frac{1}{2}^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&2\sin 37\displaystyle \frac{1}{2}^{\circ}\cos 7\displaystyle \frac{1}{2}^{\circ}\\ &=\sin \left (37\displaystyle \frac{1}{2}+7\displaystyle \frac{1}{2} \right )^{\circ}+\displaystyle \sin \left ( 37\displaystyle \frac{1}{2}-7\displaystyle \frac{1}{2} \right )^{\circ}\\ &=\sin 45^{\circ}+\sin 30^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{2}+\displaystyle \frac{1}{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{2}+1 \right )\\ \textrm{b}.\quad&2\cos 82\displaystyle \frac{1}{2}^{\circ}\sin 37\displaystyle \frac{1}{2}^{\circ}\\ &=\sin \left (82\displaystyle \frac{1}{2}+37\displaystyle \frac{1}{2} \right )^{\circ}-\displaystyle \sin \left ( 82\displaystyle \frac{1}{2}-37\displaystyle \frac{1}{2} \right )^{\circ}\\ &=\sin 120^{\circ}-\sin 45^{\circ}\\ &=\sin \left ( 180^{\circ}-60^{\circ} \right )-\sin 45^{\circ}\\ &=\sin 60^{\circ}-\sin 45^{\circ}\\ &=\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}\\ &=\displaystyle \frac{1}{2}\left ( \sqrt{3}-\sqrt{2} \right ) \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa}\\ &\textrm{a}.\quad 2\cos \left ( \displaystyle \frac{1}{4}\pi +\theta \right )\cos \left ( \displaystyle \frac{3}{4}\pi -\theta \right )=\sin 2\theta -1\\ &\textrm{b}.\quad 2\sin \left ( 315^{\circ}+B \right )\sin \left ( 45^{\circ}-B \right )=\sin 2B-1\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\textrm{a}.\quad&2\cos \left ( \displaystyle \frac{1}{4}\pi +\theta \right )\cos \left ( \displaystyle \frac{3}{4}\pi -\theta \right )\\ &=\cos \left ( \displaystyle \frac{1}{4}\pi +\theta +\displaystyle \frac{3}{4}\pi -\theta \right )+\cos \left ( \displaystyle \frac{1}{4}\pi +\theta -\left (\displaystyle \frac{3}{4}\pi -\theta \right ) \right )\\ &=\cos \left ( \displaystyle \frac{1}{4}\pi +\theta +\displaystyle \frac{3}{4}\pi -\theta \right )+\cos \left ( \displaystyle \frac{1}{4}\pi +\theta -\displaystyle \frac{3}{4}\pi +\theta \right )\\ &=\cos \pi +\cos \left ( 2\theta -\displaystyle \frac{1}{2}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2\theta \right )\\ &=-1+\sin 2\theta \qquad (\textbf{ingat sudut yang berelasi})\\ &=\color{red}\sin 2\theta -1\qquad\quad \color{black}\blacksquare\\ \textrm{b}.\quad&2\sin \left ( 315^{\circ}+B \right )\sin \left ( 45^{\circ}-B \right )\\ &=-\left (\cos \left ( 315^{\circ}+B+45^{\circ}-B \right )-\cos \left ( 315^{\circ}+B-\left ( 45^{\circ}-B \right ) \right ) \right )\\ &=-\left ( \cos 360^{\circ}-\cos \left ( 315^{\circ}-45^{\circ}+B+B \right ) \right )\\ &-=-\left ( 1-\cos \left (270^{\circ}+2B \right ) \right )\\ &=-\left (1-\sin 2B \right )\qquad (\textbf{ingat sudut yang berelasi})\\ &=\color{red}\sin 2B -1\qquad\quad \color{black}\blacksquare \end{aligned} \end{array}$
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