Lanjutan 6 Materi Rumus Jumlah dan Selisih Sinus dan Cosinus

Untuk rumus jum lah dan selisih sinus dan cosinus untuk sudut berupa X dan Y adalah sebagai berikut:

$\{\begin{array}{ll}\sin X+\sin Y& =2\sin \frac{1}{2}(X+Y)\cos \frac{1}{2}(X-Y)\\ \sin X-\sin Y& =2\cos \frac{1}{2}(X+Y)\sin \frac{1}{2}(X-Y)\\ \cos X+\cos Y& =2\cos \frac{1}{2}(X+Y)\cos \frac{1}{2}(X-Y)\\ \cos X-\cos Y& =-2\sin \frac{1}{2}(X+Y)\sin \frac{1}{2}(X-Y)\end{array}$.

Berikut contoh proses pembuktian rumus no.1, yaitu:

$\begin{array}{l}\\ \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta & & \\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta & +& \\ \sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \sin X+\sin Y=2\sin {\displaystyle \frac{(X+Y)}{2}\cos {\displaystyle \frac{(X-Y)}{2}}}\end{array}$.

Dan berikut proses pembuktian rumus no. 2, yaitu:

$\begin{array}{l}\\ \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta & & \\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta & -& \\ \sin (\alpha +\beta )-\sin (\alpha -\beta )=2\cos \alpha \sin \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \sin X-\sin Y=2\cos {\displaystyle \frac{(X+Y)}{2}\sin {\displaystyle \frac{(X-Y)}{2}}}\end{array}$

Berikut untuk proses pembktian rumus no. 3, yaitu:

$\begin{array}{l}\\ \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta & & \\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta & +& \\ \cos (\alpha +\beta )+\cos (\alpha -\beta )=2\cos \alpha \cos \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \cos X+\cos Y=2\cos {\displaystyle \frac{(X+Y)}{2}\cos {\displaystyle \frac{(X-Y)}{2}}}\end{array}$

Dan berikut tyang terakhir untuk bukti pada rumus terakhir no.4, yaitu:

$\begin{array}{l}\\ \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta & & \\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta & -& \\ \cos (\alpha +\beta )-\cos (\alpha -\beta )=-2\cos \alpha \sin \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \cos X-\cos Y=-2\sin {\displaystyle \frac{(X+Y)}{2}\sin {\displaystyle \frac{(X-Y)}{2}}}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sederhanakanlah bentuk berikut ini}\\ & \begin{array}{l}\\ \text{a}.& \sin (2x+{60}^{\circ })-\sin (2x-{60}^{\circ })\\ \text{b}.& \sin (x+{\displaystyle \frac{1}{4}m})+\sin (x-{\displaystyle \frac{1}{4}m})\\ \text{c}.& \cos (2x+4y)-\cos (2x-4y)\\ \text{d}.& \cos \left({\displaystyle \frac{5}{4}m+3x}\right)+\cos \left({\displaystyle \frac{5}{4}m-3x}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ \\ & \begin{array}{rl}\text{a}.& \sin (2x+{60}^{\circ })-\sin (2x-{60}^{\circ })\\ & =2\cos {\displaystyle \frac{(2x+{60}^{\circ })+(2x-{60}^{\circ })}{2}\sin {\displaystyle \frac{(2x+{60}^{\circ })-(2x-{60}^{\circ })}{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{array}\\ & \begin{array}{rl}\text{b}.& \sin (x+{\displaystyle \frac{1}{4}m})+\sin (x-{\displaystyle \frac{1}{4}m})\\ & =2\sin {\displaystyle \frac{(x+{\displaystyle \frac{1}{4}m+x-\frac{1}{4}m})}{2}\cos {\displaystyle \frac{(x+{\displaystyle \frac{1}{4}m-(x-\frac{1}{4}m)})}{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{array}\\ & \begin{array}{rl}\text{c}.& \cos (2x+4y)-\cos (2x-4y)\\ & =-2\sin {\displaystyle \frac{(2x+4y)+(2x-4y)}{2}\sin {\displaystyle \frac{(2x+4y)-(2x-4y)}{2}}}\\ & =-2\sin {\displaystyle \frac{4x}{2}\sin {\displaystyle \frac{8y}{2}}}\\ & =-2\sin 2x\sin 4y\end{array}\\ & \begin{array}{rl}\text{d}.& \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{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}{\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }}& ={\displaystyle \frac{2\sin {\displaystyle \frac{(3\alpha +\alpha )}{2}\cos {\displaystyle \frac{(3\alpha -\alpha )}{2}}}}{2\cos {\displaystyle \frac{(3\alpha +\alpha )}{2}\cos {\displaystyle \frac{(3\alpha -\alpha )}{2}}}}}\\ & ={\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }}\\ & =\tan 2\alpha ◼\\ & \\ & \\ & \mathbf{\text{atau}}\Rightarrow \\ & \text{mungkin lumayan rumit}\\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }}& ={\displaystyle \frac{3\sin \alpha -4{\sin }^3\alpha +\sin \alpha }{4{\cos }^3\alpha -3\cos \alpha +\cos \alpha }}\\ & ={\displaystyle \frac{4\sin \alpha -4{\sin }^3\alpha }{4{\cos }^3\alpha -2\cos \alpha }}\\ & ={\displaystyle \frac{4\sin \alpha (1-{\sin }^2\alpha )}{2\cos \alpha (2{\cos }^2\alpha -1)}={\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times {\displaystyle \frac{{\cos }^2\alpha }{\cos 2\alpha }}}}\\ & ={\displaystyle \frac{2\tan \alpha {\cos }^2\alpha }{\cos 2\alpha }}\\ & ={\displaystyle \frac{2\tan \alpha {\cos }^2\alpha }{{\cos }^2\alpha -{\sin }^2\alpha }}\\ & ={\displaystyle \frac{2\tan \alpha }{{\displaystyle \frac{{\cos }^2\alpha }{{\cos }^2\alpha }-{\displaystyle \frac{{\sin }^2\alpha }{{\cos }^2\alpha }}}}={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}}\\ & =\tan 2\alpha ◼\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }=\tan \theta }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}{\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }}& ={\displaystyle \frac{2\cos {\displaystyle \frac{(5\theta +3\theta )}{2}\sin {\displaystyle \frac{(5\theta -3\theta )}{2}}}}{2\cos {\displaystyle \frac{(3\theta +5\theta )}{2}\cos {\displaystyle \frac{(3\theta -5\theta )}{2}}}}}\\ & ={\displaystyle \frac{\sin \theta }{\cos (-\theta )}={\displaystyle \frac{\sin \theta }{\cos \theta }}}\\ & =\tan \theta ◼\\ \\ \mathbf{\text{Catatan}}:& \text{ingat bahwa saat}\cos \text{sudut}\\ & -\theta =\theta ,\text{sehingga}\cos (-\theta )=\cos \theta \end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }=\cot 2\beta }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}{\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }}& ={\displaystyle \frac{2\cos {\displaystyle \frac{(3\beta +\beta )}{2}\sin {\displaystyle \frac{(3\beta -\beta )}{2}}}}{-2\sin {\displaystyle \frac{(\beta +3\beta )}{2}\sin {\displaystyle \frac{(\beta -3\beta )}{2}}}}}\\ & ={\displaystyle \frac{\cos 2\beta \times \sin \beta }{\sin 2\beta (-\sin (-\beta ))}={\displaystyle \frac{\cos 2\beta }{\sin 2\beta }}}\\ & =\cot 2\beta ◼\\ \\ \mathbf{\text{Catatan}}:& \text{ingat bahwa saat}\sin \text{sudut}\\ & -\beta =-\beta ,\text{sehingga}\sin (-\beta )=-\sin \beta \\ & \text{dan}-\sin (-\beta )=-(-\sin \beta )=\sin \beta \end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukan nilai dari}\\ & \begin{array}{l}\\ \text{a}.& \cos {105}^{\circ }+\cos {15}^{\circ }\\ \text{b}.& \sin {105}^{\circ }-\sin {15}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{a}.& \cos {105}^{\circ }+\cos {15}^{\circ }\\ & =2\cos \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\cos \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)\\ & =2\cos {\displaystyle \frac{{120}^{\circ }}{2}\cos {\displaystyle \frac{{90}^{\circ }}{2}}}\\ & =2\cos {60}^{\circ }\cos {45}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}}\end{array}\\ & \begin{array}{rl}\text{b}.& \sin {105}^{\circ }-\sin {15}^{\circ }\\ & =2\cos \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)\\ & =2\cos {\displaystyle \frac{{120}^{\circ }}{2}\sin {\displaystyle \frac{{90}^{\circ }}{2}}}\\ & =2\cos {60}^{\circ }\sin {45}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukan nilai dari}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\cos {75}^{\circ }+\cos {15}^{\circ }}{\sin {75}^{\circ }-\sin {15}^{\circ }}}\\ \\ \text{b}.& {\displaystyle \frac{\cos {195}^{\circ }-\cos {105}^{\circ }}{\sin {105}^{\circ }-\sin {15}^{\circ }}}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{a}.& {\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 }}\\ & =\sqrt{3}\end{array}\\ & \begin{array}{rl}\text{b}.& {\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 ({180}^{\circ }-{30}^{\circ })}{\cos {60}^{\circ }}}}\\ & =-{\displaystyle \frac{\sin {30}^{\circ }}{\cos {60}^{\circ }}}\\ & =-{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =-1\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B.K. 2017. Matematika untuk Kelas SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.