Untuk rumus jum lah dan selisih sinus dan cosinus untuk sudut berupa X dan Y adalah sebagai berikut:
$\begin{cases} \sin X+\sin Y & =2\sin \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \sin X-\sin Y & =2\cos \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right )\\ \cos X+\cos Y & =2\cos \frac{1}{2}\left (X+Y \right )\cos \frac{1}{2}\left ( X-Y \right ) \\ \cos X-\cos Y & =-2\sin \frac{1}{2}\left (X+Y \right )\sin \frac{1}{2}\left ( X-Y \right ) \end{cases}$.
Berikut contoh proses pembuktian rumus no.1, yaitu:
$\begin{array}{lll}\\ \sin \left (\alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta &&\\ \sin \left (\alpha -\beta \right )=\sin \alpha \cos \beta -\cos \alpha \sin \beta &+&\\\hline \sin \left (\alpha +\beta \right )+\sin \left (\alpha -\beta \right )=2\sin \alpha \cos \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \sin X+\sin Y=2\sin \displaystyle \frac{(X+Y)}{2}\cos \displaystyle \frac{(X-Y)}{2} \end{array}$.
Dan berikut proses pembuktian rumus no. 2, yaitu:
$\begin{array}{lll}\\ \sin \left (\alpha +\beta \right )=\sin \alpha \cos \beta +\cos \alpha \sin \beta &&\\ \sin \left (\alpha -\beta \right )=\sin \alpha \cos \beta -\cos \alpha \sin \beta &-&\\\hline \sin \left (\alpha +\beta \right )-\sin \left (\alpha -\beta \right )=2\cos \alpha \sin \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \sin X-\sin Y=2\cos \displaystyle \frac{(X+Y)}{2}\sin \displaystyle \frac{(X-Y)}{2} \end{array}$
Berikut untuk proses pembktian rumus no. 3, yaitu:
$\begin{array}{lll}\\ \cos \left (\alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta &&\\ \cos \left (\alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta &+&\\\hline \cos \left (\alpha +\beta \right )+\cos \left (\alpha -\beta \right )=2\cos \alpha \cos \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \cos X+\cos Y=2\cos \displaystyle \frac{(X+Y)}{2}\cos \displaystyle \frac{(X-Y)}{2} \end{array}$
Dan berikut tyang terakhir untuk bukti pada rumus terakhir no.4, yaitu:
$\begin{array}{lll}\\ \cos \left (\alpha +\beta \right )=\cos \alpha \cos \beta -\sin \alpha \sin \beta &&\\ \cos \left (\alpha -\beta \right )=\cos \alpha \cos \beta +\sin \alpha \sin \beta &-&\\\hline \cos \left (\alpha +\beta \right )-\cos \left (\alpha -\beta \right )=-2\cos \alpha \sin \beta &&\\\\ \color{red}\textrm{saat}\quad\color{blue}X=\alpha +\beta \qquad\qquad X=\alpha +\beta &&\\ \qquad\quad Y=\alpha -\beta\quad +\qquad Y=\alpha -\beta &-&\\\hline \qquad\quad X+Y=2\alpha\qquad\quad X-Y=2\beta &&\\ \qquad\quad 2\alpha =X+Y\qquad\quad 2\beta =X-Y&&\\ \qquad\quad \alpha =\displaystyle \frac{X+Y}{2}\qquad\quad \beta =\displaystyle \frac{X-Y}{2}\\ \textbf{Sehingga rumus akan menjadi}&&\\ \cos X-\cos Y=-2\sin \displaystyle \frac{(X+Y)}{2}\sin \displaystyle \frac{(X-Y)}{2} \end{array}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Sederhanakanlah bentuk berikut ini}\\ &\begin{array}{ll}\\ \textrm{a}.&\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ \textrm{b}.&\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ \textrm{c}.&\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ \textrm{d}.&\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right ) \end{array}\\\\ &\color{blue}\textbf{Jawab}:\\\\ &\begin{aligned}\textrm{a}.\quad &\sin \left ( 2x+60^{\circ} \right )-\sin \left ( 2x-60^{\circ} \right )\\ &=2\cos \displaystyle \frac{\left (2x+60^{\circ} \right )+\left ( 2x-60^{\circ} \right )}{2}\sin \displaystyle \frac{\left (2x+60^{\circ} \right )-\left ( 2x-60^{\circ} \right )}{2}\\ &=2\cos \displaystyle \frac{4x}{2}\sin \displaystyle \frac{120^{\circ}}{2}\\ &=2\cos 2x\sin 60^{\circ}\\ &=2\cos 2x.\left ( \displaystyle \frac{1}{2}\sqrt{3} \right )\\ &=\sqrt{3}\cos 2x \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad &\sin \left ( x+\displaystyle \frac{1}{4}m \right )+\sin \left ( x-\displaystyle \frac{1}{4}m \right )\\ &=2\sin \displaystyle \frac{\left (x+\displaystyle \frac{1}{4}m+x-\frac{1}{4}m \right )}{2}\cos \displaystyle \frac{\left (x+\displaystyle \frac{1}{4}m-\left (x-\frac{1}{4}m \right ) \right )}{2}\\ &=2\sin \displaystyle \frac{2x}{2}\cos \displaystyle \frac{\displaystyle \frac{2}{4}m}{2}\\ &=2\sin x\cos \displaystyle \frac{1}{4}m \end{aligned} \\ &\begin{aligned}\textrm{c}.\quad &\cos \left ( 2x+4y \right )-\cos \left ( 2x-4y \right )\\ &=-2\sin \displaystyle \frac{\left (2x+4y \right )+\left ( 2x-4y \right )}{2}\sin \displaystyle \frac{\left (2x+4y \right )-\left ( 2x-4y \right )}{2}\\ &=-2\sin \displaystyle \frac{4x}{2}\sin \displaystyle \frac{8y}{2}\\ &=-2\sin 2x\sin 4y\\ \end{aligned} \\ &\begin{aligned}\textrm{d}.\quad &\cos \left ( \displaystyle \frac{5}{4}m+3x \right )+\cos \left ( \displaystyle \frac{5}{4}m-3x \right )\\ &=2\cos \displaystyle \frac{\left (\displaystyle \frac{5}{4}m+3x+\displaystyle \frac{5}{4}m-3x \right )}{2}\cos \displaystyle \frac{\left (\displaystyle \frac{5}{4}m+3x-\left (\displaystyle \frac{5}{4}m-3x \right ) \right )}{2}\\ &=2\cos \displaystyle \frac{5m}{2}\cos \displaystyle \frac{6x}{2}\\ &=2\cos \displaystyle \frac{5x}{2}\cos 3x\\ \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{llllllll}\\ \displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma \end{array}\\\\ &\textbf{Bukti}\\ &\begin{array}{|l|l|}\hline \begin{aligned}\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\color{red}\displaystyle \frac{2\sin \displaystyle \frac{\left (3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\alpha +\alpha \right )}{2}\cos \displaystyle \frac{\left ( 3\alpha -\alpha \right )}{2}}\\ &=\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }\\ &=\tan 2\alpha \qquad \blacksquare\\ &\\ &\\ &\quad\qquad\qquad\textbf{atau}\Rightarrow\qquad \\ &\textrm{mungkin lumayan rumit}\\ &\\ & \end{aligned}&\begin{aligned}\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }&=\displaystyle \frac{3\sin \alpha -4\sin ^{3}\alpha +\sin \alpha }{4\cos ^{3}\alpha -3\cos \alpha +\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha -4\sin ^{3}\alpha }{4\cos ^{3}\alpha -2\cos \alpha }\\ &=\displaystyle \frac{4\sin \alpha \left ( 1-\sin ^{2}\alpha \right )}{2\cos \alpha \left ( 2\cos ^{2}\alpha -1 \right )}=\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times \displaystyle \frac{\cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos 2\alpha }\\ &=\displaystyle \frac{2\tan \alpha \cos ^{2}\alpha }{\cos^{2}\alpha -\sin ^{2}\alpha }\\ &=\displaystyle \frac{2\tan \alpha }{\displaystyle \frac{\cos ^{2}\alpha }{\cos ^{2}\alpha }-\displaystyle \frac{\sin ^{2}\alpha }{\cos ^{2}\alpha }}=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ &=\tan 2\alpha \qquad \blacksquare \end{aligned} \\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{lll}\\ \displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }=\tan \theta \end{array}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta}&=\color{red}\displaystyle \frac{2\cos \displaystyle \frac{\left (5\theta +3\theta \right )}{2}\sin \displaystyle \frac{\left ( 5\theta -3\theta \right )}{2}}{2\cos \displaystyle \frac{\left ( 3\theta +5\theta \right )}{2}\cos \displaystyle \frac{\left ( 3\theta -5\theta \right )}{2}}\\ &=\displaystyle \frac{\sin \theta }{\cos (-\theta ) }=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\tan \theta \qquad \blacksquare\\\\ \textbf{Catatan}:\quad&\textrm{ingat bahwa saat}\: \: \cos\: \: \textrm{sudut}\\ &-\theta=\theta,\: \: \textrm{sehingga}\: \: \cos (-\theta )=\cos \theta \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Buktikanlah bahwa}\\ &\begin{array}{lll}\\ \displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }=\cot 2\beta \end{array}\\\\ &\textbf{Bukti}\\ &\begin{aligned}\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta}&=\color{red}\displaystyle \frac{2\cos \displaystyle \frac{\left (3\beta +\beta \right )}{2}\sin \displaystyle \frac{\left ( 3\beta -\beta \right )}{2}}{-2\sin \displaystyle \frac{\left ( \beta +3\beta \right )}{2}\sin \displaystyle \frac{\left ( \beta -3\beta \right )}{2}}\\ &=\displaystyle \frac{\cos 2\beta \times \sin \beta }{\sin 2\beta \left ( -\sin (-\beta ) \right ) }=\displaystyle \frac{\cos 2\beta }{\sin 2\beta }\\ &=\cot 2\beta \qquad \blacksquare\\\\ \textbf{Catatan}:\quad&\textrm{ingat bahwa saat}\: \: \sin\: \: \textrm{sudut}\\ &-\beta =-\beta ,\: \: \textrm{sehingga}\: \: \sin (-\beta )=-\sin \beta \\ &\textrm{dan}\: \: -\sin (-\beta )=-\left ( -\sin \beta \right )=\sin \beta \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 5.&\textrm{Tentukan nilai dari}\\ &\begin{array}{ll}\\ \textrm{a}.&\cos 105^{\circ}+\cos 15^{\circ}\\ \textrm{b}.&\sin 105^{\circ}-\sin 15^{\circ}\end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{a}.\quad&\cos 105^{\circ}+\cos 15^{\circ}\\ &=2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\cos \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{120^{\circ}}{2}\cos \displaystyle \frac{90^{\circ}}{2}\\ &=2\cos 60^{\circ}\cos 45^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\sin 105^{\circ}-\sin 15^{\circ}\\ &=2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )\\ &=2\cos \displaystyle \frac{120^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}\\ &=2\cos 60^{\circ}\sin 45^{\circ}\\ &=2\left ( \displaystyle \frac{1}{2} \right )\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\\ &=\color{red}\displaystyle \frac{1}{2}\sqrt{2} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Tentukan nilai dari}\\ &\begin{array}{ll}\\ \textrm{a}.&\displaystyle \frac{\cos 75^{\circ}+\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\\\ \textrm{b}.&\displaystyle \frac{\cos 195^{\circ}-\cos 105^{\circ}}{\sin 105^{\circ}-\sin 15^{\circ}} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{a}.\quad&\displaystyle \frac{\cos 75^{\circ}+\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{2\cos \left ( \displaystyle \frac{75^{\circ}+15^{\circ}}{2} \right )\cos \left ( \displaystyle \frac{75^{\circ}-15^{\circ}}{2} \right )}{2\cos \left ( \displaystyle \frac{75^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{75^{\circ}-15^{\circ}}{2} \right )}\\ &=\displaystyle \frac{\cos \displaystyle \frac{90^{\circ}}{2}\cos \displaystyle \frac{60^{\circ}}{2}}{\cos \displaystyle \frac{90^{\circ}}{2}\sin \displaystyle \frac{60^{\circ}}{2}}\\ &=\displaystyle \frac{\cos 30^{\circ}}{\sin 30^{\circ}}=\cot 30^{\circ}\\ &=\color{red}\sqrt{3} \end{aligned} \\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{\cos 195^{\circ}-\cos 105^{\circ}}{\sin 105^{\circ}-\sin 15^{\circ}}\\ &=\displaystyle \frac{-2\sin \left ( \displaystyle \frac{195^{\circ}+105^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{195^{\circ}-105^{\circ}}{2} \right )}{2\cos \left ( \displaystyle \frac{105^{\circ}+15^{\circ}}{2} \right )\sin \left ( \displaystyle \frac{105^{\circ}-15^{\circ}}{2} \right )}\\ &=-\displaystyle \frac{\sin \displaystyle \frac{300^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}}{\cos \displaystyle \frac{120^{\circ}}{2}\sin \displaystyle \frac{90^{\circ}}{2}}\\ &=-\displaystyle \frac{\sin 150^{\circ}}{\cos 60^{\circ}}=-\displaystyle \frac{\sin \left ( 180^{\circ}-30^{\circ} \right )}{\cos 60^{\circ}}\\ &=-\displaystyle \frac{\sin 30^{\circ}}{\cos 60^{\circ}}\\ &=-\displaystyle \frac{\displaystyle \frac{1}{2}}{\displaystyle \frac{1}{2}}\\ &=\color{red}-1 \end{aligned} \end{array}$.
DAFTAR PUSTAKA
- Noormandiri, B.K. 2017. Matematika untuk Kelas SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
- Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.
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