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.

Lanjutan 5 Materi Rumus Perkalian Sinus dan Cosinus

C.   Rumus Perkalian Sinus dan Cosinus

Untuk sudut  $\alpha \text{dan}\beta$ berlaku rumus-rumus

$\begin{array}{rl}& \{\begin{array}{ll}2\sin \alpha \cos \gamma & =\sin (\alpha +\gamma )+\sin (\alpha -\gamma )\\ 2\cos \alpha \sin \gamma & =\sin (\alpha +\gamma )-\sin (\alpha -\gamma )\end{array}\\ & \{\begin{array}{ll}2\cos \alpha \cos \gamma & =\cos (\alpha +\gamma )+\cos (\alpha -\gamma )\\ -2\sin \alpha \sin \gamma & =\cos (\alpha +\gamma )-\cos (\alpha -\gamma )\end{array}\end{array}$.

Sebagai buktinya akan ditunjukkan pada tulisan berikut

Bukti untuk nomor pertama

$\begin{array}{rl}\sin & (\alpha +\beta )+\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & =2\sin \alpha \cos \beta \end{array}$.

Selanjutnya untuk bukti baris kedua yaitu:

$\begin{array}{rl}\sin & (\alpha +\beta )-\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & -(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\\ & =2\cos \alpha \sin \beta \end{array}$.

Dan bukti untuk rumus ketiga

$\begin{array}{rl}\cos & (\alpha +\beta )+\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & +\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ & =2\cos \alpha \cos \beta \end{array}$.

Adapun bukti untuk rumus yang tertakhir adalah:

$\begin{array}{rl}\cos & (\alpha +\beta )-\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & -(\cos \alpha \cos \beta +\sin \alpha \sin \beta )\\ & =-2\sin \alpha \sin \beta \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {80}^{\circ }\sin {50}^{\circ }\\ & \text{b}.2\sin {80}^{\circ }\cos {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {80}^{\circ }\sin {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })-\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }-\sin {30}^{\circ }\\ & =\sin {130}^{\circ }-{\displaystyle \frac{1}{2}}\\ \text{b}.& 2\sin {80}^{\circ }\cos {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })+\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }+\sin {30}^{\circ }\\ & =\sin {130}^{\circ }+{\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {75}^{\circ }\cos {15}^{\circ }\\ & \text{b}.\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {75}^{\circ }\cos {15}^{\circ }\\ & =\cos ({75}^{\circ }+{15}^{\circ })+\cos ({75}^{\circ }-{15}^{\circ })\\ & =\cos {90}^{\circ }+\cos {60}^{\circ }\\ & =0+{\displaystyle \frac{1}{2}={\displaystyle \frac{1}{2}}}\\ \text{b}.& \sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ & =-{\displaystyle \frac{1}{2}(2\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}(\cos \left({\displaystyle \frac{3\pi }{8}+\frac{\pi }{8}}\right)-\cos \left({\displaystyle \frac{3\pi }{8}-\frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{4\pi }{8}-\cos {\displaystyle \frac{2\pi }{8}}}\right)}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{1}{2}\pi -\cos {\displaystyle \frac{1}{4}\pi }}\right)}\\ & =-{\displaystyle \frac{1}{2}(0-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & =-{\displaystyle \frac{1}{2}(-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & ={\displaystyle \frac{1}{4}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari bentuk berikut}\\ \\ & \text{a}.2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \text{b}.2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(37{\displaystyle \frac{1}{2}+7{\displaystyle \frac{1}{2}}}\right)}^{\circ }+{\displaystyle \sin {\left(37{\displaystyle \frac{1}{2}-7{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {45}^{\circ }+\sin {30}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{2}+1)}\\ \text{b}.& 2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(82{\displaystyle \frac{1}{2}+37{\displaystyle \frac{1}{2}}}\right)}^{\circ }-{\displaystyle \sin {\left(82{\displaystyle \frac{1}{2}-37{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {120}^{\circ }-\sin {45}^{\circ }\\ & =\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ }\\ & =\sin {60}^{\circ }-\sin {45}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{3}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & \text{a}.2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)=\sin 2\theta -1\\ & \text{b}.2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)=\sin 2B-1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& 2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -\left({\displaystyle \frac{3}{4}\pi -\theta }\right)}\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -{\displaystyle \frac{3}{4}\pi +\theta }}\right)\\ & =\cos \pi +\cos (2\theta -{\displaystyle \frac{1}{2}\pi })\\ & =-1+\cos \left({\displaystyle \frac{1}{2}\pi -2\theta }\right)\\ & =-1+\sin 2\theta (\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2\theta -1◼\\ \text{b}.& 2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)\\ & =-(\cos ({315}^{\circ }+B+{45}^{\circ }-B)-\cos ({315}^{\circ }+B-({45}^{\circ }-B)))\\ & =-(\cos {360}^{\circ }-\cos ({315}^{\circ }-{45}^{\circ }+B+B))\\ & -=-(1-\cos ({270}^{\circ }+2B))\\ & =-(1-\sin 2B)(\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2B-1◼\end{array}\end{array}$

Lanjutan 4 Materi Rumus Sudut Paruh pada Trigonometri

Sebelumnya telah dituliskan bahwa

$\begin{array}{l}\\ \bullet & \sin \alpha =2\sin {\displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha }\\ \bullet & \cos \alpha ={\cos }^2{\displaystyle \frac{1}{2}\alpha -{\sin }^2{\displaystyle \frac{1}{2}\alpha }}\\ \bullet & \tan \alpha ={\displaystyle \frac{2\tan {\displaystyle \frac{1}{2}\alpha }}{1-{\tan }^2{\displaystyle \frac{1}{2}\alpha }}}\end{array}$.

 B. 2. 1   Rumus Sudut Paruh untuk Sinus

Perhatikanlah Identitas trigonometri berikut

Perhatikanlah ilsutrasi segitiga ABC berikut

• ${\sin }^2\alpha +{\cos }^2\alpha =1$
• ${\tan }^2\alpha -{\sec }^2\alpha =-1$
• ${\cot }^2\alpha -{\csc }^2\alpha =-1$

Serta pada pembahasan sebelumnya untuk sudut rangkap bahwa

• $\sin 2\alpha =2\sin \alpha \cos \alpha$
• $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, dan 
• $\cos 2\alpha =2{\cos }^2\alpha -1$, serta
• $\cos 2\alpha =1-2{\sin }^2\alpha$

Maka rumus $\sin {\displaystyle \frac{1}{2}\alpha }$ adalah:

$\begin{array}{rl}\cos 2\alpha & =1-2{\sin }^2\alpha \\ 2{\sin }^2\alpha & =1-\cos 2\alpha \\ {\sin }^2\alpha & ={\displaystyle \frac{1-\cos 2\alpha }{2}}\\ \sin \alpha & =\sqrt{{\displaystyle \frac{1-\cos 2\alpha }{2}}}\\ \text{saat}& \text{sudut}\alpha \text{diganti}{\displaystyle \frac{1}{2}\alpha ,}\\ \text{maka}& \text{rumus akan menjadi}\\ \sin \left({\displaystyle \frac{1}{2}\alpha }\right)& =\sqrt{{\displaystyle \frac{1-\cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)}{2}}}\\ \sin {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{2}}}\end{array}$.

B. 2. 2   Rumus Sudut Paruh untuk Cosinus

Sedangkan untuk rumus cosinusnya adalah:

$\begin{array}{rl}\cos 2\alpha & =2{\cos }^2\alpha -1\\ 2{\cos }^2\alpha & =1+\cos 2\alpha \\ {\cos }^2\alpha & ={\displaystyle \frac{1+\cos 2\alpha }{2}}\\ \cos \alpha & =\sqrt{{\displaystyle \frac{1+\cos 2\alpha }{2}}}\\ \text{saat}& \text{sudut}\alpha \text{diganti}{\displaystyle \frac{1}{2}\alpha ,}\\ \text{maka}& \text{rumus akan menjadi}\\ \cos \left({\displaystyle \frac{1}{2}\alpha }\right)& =\sqrt{{\displaystyle \frac{1+\cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)}{2}}}\\ \cos {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1+\cos \alpha }{2}}}\end{array}$

B. 2. 3   Rumus Sudut Paruh untuk Tangen

Adapun untuk rumus tangen adalah:

$\begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle \frac{\sin {\displaystyle \frac{1}{2}\alpha }}{\cos {\displaystyle \frac{1}{2}\alpha }}}\\ & ={\displaystyle \frac{\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{2}}}}{\pm \sqrt{{\displaystyle \frac{1+\cos \alpha }{2}}}}}\\ & =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }}}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }}},\text{maka}\\ & =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }\times {\displaystyle \frac{1-\cos \alpha }{1-\cos \alpha }}}}\\ & =\pm \sqrt{{\displaystyle \frac{{(1-\cos \alpha )}^2}{1-{\cos }^2\alpha }}}\\ & =\pm \sqrt{{\displaystyle \frac{{(1-\cos \alpha )}^2}{{\sin }^2\alpha }}}\\ & ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{\sin \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{\sin \alpha }{1+\cos \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{\sin \alpha },\text{maka}}}\\ & ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\times \frac{1+\cos \alpha }{1+\cos \alpha }}\\ & ={\displaystyle \frac{1-{\cos }^2\alpha }{\sin \alpha (1+\cos \alpha )}}\\ & ={\displaystyle \frac{{\sin }^2\alpha }{\sin \alpha (1+\cos \alpha )}}\\ & ={\displaystyle {\displaystyle \frac{\sin \alpha }{1+\cos \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa nilai}{\cos }^2{\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}{\cos }^2{\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1+\cos \alpha }{2},\text{maka}}}\\ & ={\displaystyle {\displaystyle \frac{1+\cos \alpha }{2}\times {\displaystyle \frac{{\displaystyle \frac{1}{\cos \alpha }}}{{\displaystyle \frac{1}{\cos \alpha }}}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{\cos \alpha }+{\displaystyle \frac{\cos \alpha }{\cos \alpha }}}}{{\displaystyle \frac{2}{\cos \alpha }}}}\\ & ={\displaystyle \frac{\sec \alpha +1}{2\sec \alpha }}\\ & ={\displaystyle {\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa nilai}\sin {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}{\sin }^2{\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{2},\text{maka}}}\\ & ={\displaystyle \frac{1-\sqrt{{\cos }^2\alpha }}{2}}\\ & ={\displaystyle \frac{1-\sqrt{1-{\sin }^2\alpha }}{2}}\\ & ={\displaystyle \frac{1-\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{2}}\\ & ={\displaystyle \frac{1-\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{2}}\\ & ={\displaystyle \frac{2+\sin \alpha -\sin \alpha -2\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{4}}\\ & ={\displaystyle \frac{1+\sin \alpha +1-\sin \alpha -2\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{4}}\\ & ={\displaystyle \frac{{(\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha })}^2}{4}}\\ & ={\left({\displaystyle \frac{(\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha })}{2}}\right)}^2\\ \sin {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Dengan menggunakan rumus sudut}\\ & \text{paruh, tentukanlah nilai dari}\\ & \text{a}.\cos {\displaystyle \frac{\pi }{8},\text{b}.\sin {15}^{\circ }\text{c}.\tan {15}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat baik sudut}{\displaystyle \frac{\pi }{8}\text{atau}{\displaystyle \frac{\pi }{4}\text{di kuadran I}}}\\ & \text{maka}\cos {\displaystyle \frac{\pi }{8}\text{bertanda positif, sehingga}}\\ & \begin{array}{rl}\text{a}.\cos {\displaystyle \frac{\pi }{8}}& =\sqrt{{\displaystyle \frac{1+\cos {\displaystyle \frac{\pi }{4}}}{2}}}=\sqrt{{\displaystyle \frac{1+{\displaystyle \frac{1}{2}\sqrt{2}}}{2}}}\\ & =\sqrt{{\displaystyle \frac{2+\sqrt{2}}{4}}}={\displaystyle \frac{1}{2}\sqrt{2+\sqrt{2}}}\end{array}\\ & \text{Adapun untuk sudut}{15}^{\circ }\text{ataupun}{30}^{\circ }\\ & \text{akan berada di kuadran I akibatnya }\\ & \text{tandanya positif, sehingga}\\ & \begin{array}{rl}\text{b}.\sin {15}^{\circ }& =\sqrt{{\displaystyle \frac{1-\cos {30}^{\circ }}{2}}}\\ & =\sqrt{{\displaystyle \frac{1-{\displaystyle \frac{1}{2}\sqrt{3}}}{2}}}=\sqrt{{\displaystyle \frac{2-\sqrt{3}}{4}}}\\ & ={\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}\end{array}\\ & \text{Dan untuk sudut}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }}}\\ & \begin{array}{rl}\text{c}.\tan {15}^{\circ }& ={\displaystyle \frac{1-\cos {30}^{\circ }}{\sin {30}^{\circ }}}\\ & ={\displaystyle \frac{1-{\displaystyle \frac{1}{2}\sqrt{3}}}{{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{(2-\sqrt{3})}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =2-\sqrt{3}\end{array}\end{array}$.

DAFTAR PUSTAKA

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

Lanjutan 3 Materi Rumus Sudut Ganda(Rangkap) dan Sudut Paruh Pada Trigonometri

B. 1   Rumus Sudut Rangkap

Pada materi sebelumnya sudah dibahas tentang hal ini yaitu,

$\begin{array}{l}\\ \bullet & \sin 2\alpha =2\sin \alpha \cos \alpha \\ \bullet & \cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha \\ \bullet & \tan 2\alpha ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\end{array}$.

B. 2   Rumus Sudut Rangkap Tiga

Berikut untuk rumus sudut rangkap tiga:

$\begin{array}{rl}\bullet \sin 3\alpha & =3\sin \alpha -4{\sin }^3\alpha \\ \bullet \cos 3\alpha & =4{\cos }^3\alpha -3\cos \alpha \\ \bullet \tan 3\alpha & ={\displaystyle \frac{3\tan x-{\tan }^3\alpha }{1-3{\tan }^2\alpha }}\end{array}$

B. 3   Rumus Sudut Paruh

Demikian juga untuk sudut paruhnya, juga sudah dibahas pada materi sebelumnya yang mana sebelumnya sudut paruh itu diperoleh dari aplikasi rumus jum lah dan selisih dua sudut ketika sudutnya berupa setengan sudut dan sama pula, yaitu:

$\begin{array}{l}\\ \bullet & \sin \alpha =2\sin {\displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha }\\ \bullet & \cos \alpha ={\cos }^2{\displaystyle \frac{1}{2}\alpha -{\sin }^2{\displaystyle \frac{1}{2}\alpha }}\\ \bullet & \tan \alpha ={\displaystyle \frac{2\tan {\displaystyle \frac{1}{2}\alpha }}{1-{\tan }^2{\displaystyle \frac{1}{2}\alpha }}}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\sin A={\displaystyle \frac{4}{5},}\\ & \text{untuk}A\text{sudut tumpul. Tentukanlah}\\ & \text{a}.\sin 2A\\ & \text{b}.\cos 2A\\ & \text{c}.\tan 2A\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan menggunakan identitas}\\ & {\sin }^2\alpha +{\cos }^2\alpha =1,\text{dapat diperoleh}\\ & \text{nilai}\cos A,\text{yaitu}:\\ & {\cos }^2\alpha =1-{\sin }^{2}A=1-{\left({\displaystyle \frac{4}{5}}\right)}^2\\ & =1-{\displaystyle \frac{16}{25}=\frac{9}{25}\Rightarrow \cos A=\pm {\displaystyle \frac{3}{5}}}\\ & \text{ambil}\cos A=-{\displaystyle \frac{3}{5},\text{karena di kuadran II}}\\ & \text{ingat sudut tumpul}=\text{sudut di kuadran II}\\ & \begin{array}{rl}\text{a}.\sin 2A& =2\sin A\cos A\\ & =2\left({\displaystyle \frac{4}{5}}\right)(-{\displaystyle \frac{3}{5}})\\ & =-{\displaystyle \frac{24}{25}}\\ \text{b}.\cos 2A& ={\cos }^{2}A-{\sin }^{2}A\\ & ={(-{\displaystyle \frac{3}{5}})}^2-{\left({\displaystyle \frac{4}{5}}\right)}^2\\ & ={\displaystyle \frac{9}{25}-\frac{16}{25}=-\frac{7}{25}}\\ \text{c}.\tan 2A& ={\displaystyle \frac{\sin 2A}{\cos 2A}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{24}{25}}}{-{\displaystyle \frac{7}{25}}}={\displaystyle \frac{24}{7}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai dari}\\ & \text{a}.\sin {15}^{\circ }\\ & \text{b}.\cos {15}^{\circ }\\ & \text{c}.\tan {15}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Banyak cara yang bisa digunakan}.\\ & \text{berikut salah satu cara itu, yaitu}\\ & \begin{array}{rlr}\text{a.}\cos \left(2\alpha \right)& =1-2{\sin }^2\alpha ,& \text{pilih untuk}\alpha ={15}^0\\ \cos (2\times {15}^0)& =1-2{\sin }^2{15}^0\\ \cos {30}^0& =1-2{\sin }^2{15}^0\\ {\displaystyle \frac{1}{2}\sqrt{3}}& =1-2{\sin }^2{15}^0\\ \sqrt{3}& =2-4{\sin }^2{15}^0\\ {\sin }^2{15}^0& ={\displaystyle \frac{2-\sqrt{3}}{4}}\\ \sin {15}^0& ={\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}\\ \text{b.}{\sin }^2\alpha +& {\cos }^2\alpha =1,& \text{rumus identitas}\\ {\sin }^2{15}^0+& {\cos }^2{15}^0=1\\ {\cos }^2{15}^0& =1-{\sin }^2{15}^0\\ & ={\displaystyle 1-\left(\frac{2-\sqrt{3}}{4}\right),}& \text{lihat jawaban poin a)}\\ & ={\displaystyle \frac{2+\sqrt{3}}{4}}\\ \cos {15}^0& ={\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}\\ \text{c.}\tan \alpha & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ \tan {15}^0& ={\displaystyle \frac{\sin {15}^0}{\cos {15}^0}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}}{{\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}}}\\ & =\sqrt{{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}}\times \sqrt{\frac{2-\sqrt{3}}{2-\sqrt{3}}},& \text{sekawan dari penyebut}\\ & =\sqrt{{\displaystyle \frac{4-2.2\sqrt{3}+3}{4-3}}}\\ & =\sqrt{7-4\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa nilai}\\ & {\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }=\tan \alpha }\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}{\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }}& ={\displaystyle \frac{1-({\cos }^2\alpha -{\sin }^2\alpha )}{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{1-((1-{\sin }^2\alpha )-{\sin }^2\alpha )}{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{1-1+{\sin }^2\alpha +{\sin }^2\alpha }{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{2{\sin }^2\alpha }{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ & ={\displaystyle \tan \alpha ◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa nilai}\\ & \text{a}.\cos 3\alpha =4{\cos }^3\alpha -3\cos \alpha \\ & \text{b}.\tan 3\alpha ={\displaystyle \frac{3\tan x-{\tan }^3\alpha }{1-3{\tan }^2\alpha }}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.\cos 3\alpha & =\cos (2\alpha +\alpha )\\ & =\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ & =(2{\cos }^2\alpha -1)\cos \alpha -(2\sin \alpha \cos \alpha )\sin \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2{\sin }^2\alpha \cos \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2(1-{\cos }^2\alpha )\cos \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2\cos \alpha +2{\cos }^2\alpha \\ & =4{\cos }^3\alpha -3\cos \alpha ◼\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{b}.\tan 3\alpha & =\tan (2\alpha +\alpha )\\ & ={\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\right)+\tan \alpha }{1-\left({\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\right).\tan \alpha }}\\ & ={\displaystyle \frac{{\displaystyle \frac{2\tan \alpha +\tan \alpha -{\tan }^3\alpha }{1-{\tan }^2\alpha }}}{{\displaystyle \frac{(1-{\tan }^2\alpha )-2{\tan }^2\alpha }{1-{\tan }^2\alpha }}}}\\ & ={\displaystyle \frac{3\tan \alpha -{\tan }^3\alpha }{1-3{\tan }^2\alpha }◼}\end{array}\\ \hline\end{array}\end{array}$

Lanjutan 2 Materi Rumus-Rumus Trigonometri

 A. 3  Rumus  $\tan (\alpha +\beta )$ dan $\tan (\alpha -\beta )$.

Sebelumnya telah dibahas pada materi sebelumnya dan disertai pula dengan contoh soal rumus jumlah dan selisih dua sudut untuk sinus dan cosinus, yaitu:

Jika dua sudut yang dimaksud misalkan alfa dan beta, maka

$\{\begin{array}{ll}\sin (\alpha +\gamma )& =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \sin (\alpha -\gamma )& =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos (\alpha +\gamma )& =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \cos (\alpha +\gamma )& =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \end{array}$.

Sengan menggunakan fakta yang ada-rumus yang telah diketahui-kita akan terbantu dalam menemukan rumus untuk tangen, yaitu:

$\begin{array}{rl}\tan (\alpha +\beta )& ={\displaystyle \frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}}\\ & ={\displaystyle \frac{(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\times \left({\displaystyle \frac{1}{\cos \alpha \cos \beta }}\right)}{(\cos \alpha \cos \beta -\sin \alpha \sin \beta )\times \left({\displaystyle \frac{1}{\cos \alpha \cos \beta }}\right)}}\\ & ={\displaystyle \frac{{\displaystyle \frac{\sin \alpha }{\cos \alpha }+{\displaystyle \frac{\sin \beta }{\cos \beta }}}}{1-{\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}}}\\ & ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }.◼}\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\tan (\alpha +\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}\tan (\alpha +\beta )& ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}\\ & \text{dengan mengganti}\beta =-\beta \text{maka,}\\ & ={\displaystyle \frac{\tan \alpha +\tan (-\beta )}{1-\tan \alpha \tan (-\beta )}}\\ \tan (\alpha -\beta )& ={\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }.◼}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\tan {60}^{\circ }=\sqrt{3}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\tan {60}^{\circ }& =\tan ({30}^{\circ }+{30}^{\circ })\\ & ={\displaystyle \frac{\tan {30}^{\circ }+\tan {30}^{\circ }}{1-\tan {30}^{\circ }\tan {30}^{\circ }}}\\ & ={\displaystyle \frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}}\\ & ={\displaystyle \frac{2\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)}{1-{\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)}^2}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{1-{\displaystyle \frac{1}{9}\sqrt{9}}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{1-{\displaystyle \frac{3}{9}}}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{{\displaystyle \frac{6}{9}}}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{{\displaystyle \frac{2}{3}}}}}}\\ & ={\displaystyle \sqrt{3}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\\ & \tan {90}^{\circ }=\text{Tidak Terdefinisi}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\tan {90}^{\circ }& =\tan ({45}^{\circ }+{45}^{\circ })\\ & ={\displaystyle \frac{\tan {45}^{\circ }+\tan {45}^{\circ }}{1-\tan {45}^{\circ }\tan {45}^{\circ }}}\\ & ={\displaystyle \frac{2\tan {45}^{\circ }}{1-{\tan }^2{45}^{\circ }}}\\ & ={\displaystyle \frac{2\left(1\right)}{1-{\left(1\right)}^2}={\displaystyle \frac{{\displaystyle 2}}{1-1}}}\\ & ={\displaystyle \frac{2}{0}}\\ & =\text{Tidak Terdefinisi}◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\tan {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {75}^{\circ }& =\tan ({30}^{\circ }+{45}^{\circ })\\ & ={\displaystyle \frac{\tan {30}^{\circ }+\tan {45}^{\circ }}{1-\tan {30}^{\circ }\tan {45}^{\circ }}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}\sqrt{3}+1}}{1-\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)\left(1\right)}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{3}\sqrt{3}}}{1-{\displaystyle \frac{1}{3}\sqrt{3}}}={\displaystyle \frac{{\displaystyle \frac{1}{3}(3+\sqrt{3})}}{{\displaystyle \frac{1}{3}(3-\sqrt{3})}}}}\\ & ={\displaystyle \frac{3+\sqrt{3}}{3-\sqrt{3}}}\\ & ={\displaystyle \frac{3+\sqrt{3}}{3-\sqrt{3}}\times {\displaystyle \frac{3+\sqrt{3}}{3+\sqrt{3}}}}\\ & ={\displaystyle \frac{3^2+3\sqrt{3}+3\sqrt{3}+\sqrt{9}}{3^2-{\left(\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{12+6\sqrt{3}}{9-3}={\displaystyle \frac{12+6\sqrt{3}}{6}}}\\ & ={\displaystyle \frac{6}{6}(2+\sqrt{3})}\\ & =2+\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\tan {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {105}^{\circ }& =\tan ({45}^{\circ }+{60}^{\circ })\\ & ={\displaystyle \frac{\tan {45}^{\circ }+\tan {60}^{\circ }}{1-\tan {45}^{\circ }\tan {60}^{\circ }}}\\ & ={\displaystyle \frac{1+\sqrt{3}+1}{1-1.\left(\sqrt{3}\right)}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\times {\displaystyle \frac{1+\sqrt{3}}{1+\sqrt{3}}}}\\ & ={\displaystyle \frac{1^2+\sqrt{3}+\sqrt{3}+{\left(\sqrt{3}\right)}^2}{1^2-{\left(\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{4+2\sqrt{3}}{1-3}={\displaystyle \frac{4+2\sqrt{3}}{-2}}}\\ & =-2-\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\tan ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan ({270}^{\circ }+A)& ={\displaystyle \frac{\tan {270}^{\circ }+\tan A}{1-\tan {270}^{\circ }\tan A}}\\ & ={\displaystyle \frac{\mathbf{\text{TD}}+\tan A}{1-\mathbf{\text{TD}}\tan A}}\\ & \text{dengan}\mathbf{\text{TD}}\text{adalah Tidak Terdefinisi}\\ & =\text{Bentuk yang harus dihindari}\\ \text{maka gunakan}& \text{bentuk berikut ini}\\ \tan ({270}^{\circ }+A)& ={\displaystyle \frac{\sin ({270}^{\circ }+A)}{\cos ({270}^{\circ }+A)}}\\ & ={\displaystyle \frac{-\cos A}{\sin A}}\\ & =-\cot A\\ \\ \mathbf{\text{Catatan}}:& \text{lihatlah materi sebelumnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\tan ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan ({270}^{\circ }-A)& ={\displaystyle \frac{\sin ({270}^{\circ }-A)}{\cos ({270}^{\circ }-A)}}\\ & ={\displaystyle \frac{-\cos A}{-\sin A}}\\ & =\cot A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \tan \left(2\alpha \right)={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\\ \text{b}.& \tan \alpha ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}}\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\tan (\alpha +\beta )& ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta },}& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \tan (\alpha +\alpha )& ={\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha }}\\ \tan 2\alpha & ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }\alpha ◼}\\ \text{b.}\tan 2\alpha & ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha },}& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \tan 2\left({\displaystyle \frac{1}{2}\alpha }\right)& ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-{\tan }^2\left({\displaystyle \frac{1}{2}\alpha }\right)}}\\ \tan \alpha & ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-{\tan }^2\left({\displaystyle \frac{1}{2}\alpha }\right)}◼}\end{array}\end{array}$.

Lanjutan 1 Materi Rumus-Rumus Trigonometri

A. 2  Rumus  $\cos (\alpha +\beta )$ dan $\cos (\alpha -\beta )$.

Dalam penentuan rumus $\cos (\alpha -\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Jika diurai gambar di atas adalah 

Mungkin gambarnya ada tang kurang jelas, mari kita perjelas lagi gambar di atas

Perhatikan bahwa

dan

Sehingga

$\begin{array}{rl}\left[ABC\right]& =\left[ACD\right]+\left[BCD\right]\\ {\displaystyle \frac{1}{2}ab\sin ({90}^0-\alpha +\beta )}& ={\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +{\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta }}\\ \sin ({90}^0-(\alpha -\beta ))& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos (\alpha -\beta )& =\cos \alpha \cos \beta +\sin \alpha \sin \beta .◼\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\cos (\alpha +\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}& \cos (\alpha -(-\beta ))=\cos (\alpha +\beta )\\ & =\cos \alpha \cos (-\beta )+\sin \alpha \sin (-\beta )\\ & =\cos \alpha \cos \beta +\sin \alpha (-\sin \beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\cos {60}^{\circ }& =\cos ({30}^{\circ }+{30}^{\circ })\\ & =\cos {30}^{\circ }\cos {30}^{\circ }-\sin {30}^{\circ }\sin {30}^{\circ }\\ & ={\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }\\ & ={\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{1}{4}\sqrt{9}-\frac{1}{4}={\displaystyle \frac{3}{4}-\frac{1}{4}={\displaystyle \frac{2}{4}}}}\\ & ={\displaystyle \frac{1}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\cos {90}^{\circ }=0\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \cos {90}^{\circ }& =\cos ({60}^{\circ }+{30}^{\circ })\\ & =\cos {60}^{\circ }\cos {30}^{\circ }-\sin {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{3}-{\displaystyle \frac{1}{4}\sqrt{3}}}\\ & ={\displaystyle 0◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \cos {90}^{\circ }& =\cos ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \cos {90}^{\circ }& =\cos ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\cos {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {75}^{\circ }& =\cos ({30}^{\circ }+{45}^{\circ })\\ & =\cos {30}^{\circ }\cos {45}^{\circ }-\sin {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)-\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{6}-{\displaystyle \frac{1}{4}\sqrt{2}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{6}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\cos {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {105}^{\circ }& =\cos ({45}^{\circ }+{60}^{\circ })\\ & =\cos {45}^{\circ }\cos {60}^{\circ }-\sin {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}-{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}-\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }+A)& =\cos {270}^{\circ }\cos A-\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A-(-1)\sin A\\ & =0+\sin A\\ & =\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }-A)& =\cos {270}^{\circ }\cos A+\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A+(-1)\sin A\\ & =0-\sin A\\ & =-\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \cos \left(2\alpha \right)={\cos }^2\alpha -{\sin }^2\alpha \\ \text{b}.& \cos \alpha ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\cos (\alpha +\beta )& =\cos \alpha \cos \beta -\sin \alpha \sin \beta ,& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \cos (\alpha +\alpha )& =\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha ◼\\ \text{b.}\cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha ,& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)& ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)\\ \cos \alpha & ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)◼\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.

Rumus-Rumus Trigonometri

Sebelumnya kita buka arsip lama di blog ini, yaitu:

 Materi Lawas

A. Rumus Jumlah Dan selisih Sudut

A. 1  Rumus  $\sin (\alpha +\beta )$ dan $\sin (\alpha -\beta )$.

Dalam penentuan rumus $\sin (\alpha +\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Bukti:

$\begin{array}{rl}{\displaystyle \frac{AC}{\sin {90}^0}}& ={\displaystyle \frac{C{A}^{\prime }}{\sin \alpha }={\displaystyle \frac{A{A}^{\prime }}{\sin \mathrm{\angle }C}}}\\ A{A}^{\prime }& =AC.\sin \mathrm{\angle }C\\ & =AC.\sin ({90}^0-\alpha )\\ & =AC.\cos \alpha \\ & \text{dengan cara yang kurang lebih }\\ & \text{sama akan diperoleh juga}\\ A{A}^{\prime }& =AB.\cos \beta \\ & \text{selanjutnya kita tentukan luasnya, yaitu}\\ \left[ABC\right]& =\left[A{A}^{\prime }C\right]+\left[A{A}^{\prime }B\right]\\ {\displaystyle \frac{1}{2}.AB.AC.\sin (\alpha +\beta )}& ={\displaystyle \frac{1}{2}.AC.A{A}^{\prime }.\sin \alpha +{\displaystyle \frac{1}{2}.AB.A{A}^{\prime }.\sin \beta }}\\ \sin (\alpha +\beta )& ={\displaystyle \frac{AC.A{A}^{\prime }.\sin \alpha }{AB.AC}+{\displaystyle \frac{AB.A{A}^{\prime }.\sin \beta }{AB.AC}}}\\ & ={\displaystyle \frac{A{A}^{\prime }}{AB}.\sin \alpha +{\displaystyle \frac{A{A}^{\prime }}{AC}.\sin \beta }}\\ & ={\displaystyle \frac{(AB.\cos \beta )}{AB}.\sin \alpha +{\displaystyle \frac{(AC.\cos \alpha )}{AC}.\sin \beta }}\\ \sin (\alpha +\beta )& =\sin \alpha .\cos \beta +\cos \alpha .\sin \beta ◼\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\sin (\alpha -\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}& \sin (\alpha +(-\beta ))=\sin (\alpha -\beta )\\ & =\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha (-\sin \beta )\\ & =\sin \alpha \cos \beta -\cos \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\sin {60}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\sin {60}^{\circ }& =\sin ({30}^{\circ }+{30}^{\circ })\\ & =\sin {30}^{\circ }\cos {30}^{\circ }+\cos {30}^{\circ }\sin {30}^{\circ }\\ & =\sin {30}^{\circ }\cos {30}^{\circ }+\sin {30}^{\circ }\cos {30}^{\circ }\\ & =2\sin {30}^{\circ }\cos {30}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{3}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\sin {90}^{\circ }=1\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \sin {90}^{\circ }& =\sin ({60}^{\circ }+{30}^{\circ })\\ & =\sin {60}^{\circ }\cos {30}^{\circ }+\cos {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)+\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{9}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{3}{4}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{4}{4}}\\ & ={\displaystyle 1◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \sin {90}^{\circ }& =\sin ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \sin {90}^{\circ }& =\sin ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\sin {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {75}^{\circ }& =\sin ({30}^{\circ }+{45}^{\circ })\\ & =\sin {30}^{\circ }\cos {45}^{\circ }+\cos {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\sin {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {105}^{\circ }& =\sin ({45}^{\circ }+{60}^{\circ })\\ & =\sin {45}^{\circ }\cos {60}^{\circ }+\cos {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }+A)& =\sin {270}^{\circ }\cos A+\cos {270}^{\circ }\sin A\\ & =(-1)\cos A+\left(0\right)\sin A\\ & =-\cos A+0\\ & =-\cos A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }-A)& =\sin {270}^{\circ }\cos A-\cos {270}^{\circ }\sin A\\ & =(-1)\cos A-\left(0\right)\sin A\\ & =-\cos A-0\\ & =-\cos A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \sin \left(2\alpha \right)=2\sin \alpha \cos \alpha \\ \text{b}.& \sin \alpha =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right).\cos \left({\displaystyle \frac{1}{2}\alpha }\right)\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\sin (\alpha +\beta )& =\sin \alpha \cos \beta +\cos \alpha \sin \beta ,& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \sin (\alpha +\alpha )& =\sin \alpha \cos \alpha +\cos \alpha \sin \alpha \\ \sin 2\alpha & =2\sin \alpha \cos \alpha ◼\\ \text{b.}\sin 2\alpha & =2\sin \alpha \cos \alpha ,& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \sin 2\left({\displaystyle \frac{1}{2}\alpha }\right)& =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right)\cos \left({\displaystyle \frac{1}{2}\alpha }\right)\\ \sin \alpha & =2\sin \left({\displaystyle \frac{1}{2}\alpha }\right)\cos \left({\displaystyle \frac{1}{2}\alpha }\right)◼\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.

Lanjutan 3 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 16.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{75}^{\circ },{285}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{75}^{\circ },{343}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{75}^{\circ },{344}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{75}^{\circ },{345}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{75}^{\circ },{346}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}}\\ & \cos (x-30{)}^{\circ }=\cos {45}^{\circ }\\ & \iff x-{30}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={75}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{15}^{\circ }(tm)\\ & k=1\Rightarrow x_1={75}^{\circ }+{360}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{15}^{\circ }+{360}^{\circ }={345}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {75}^{\circ },{345}^{\circ }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }<x<{360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{100}^{\circ },{330}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{30}^{\circ },{330}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{120}^{\circ },{300}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{60}^{\circ },{120}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{50}^{\circ },{300}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos x^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ & \cos x^{\circ }=\cos {30}^{\circ }\\ & \iff x=\pm {30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={30}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{30}^{\circ }(tm)\\ & k=1\Rightarrow x_1={30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{30}^{\circ }+{360}^{\circ }={330}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {30}^{\circ },{330}^{\circ }}\right\}\end{array}\end{array}$

Lanjutan 2 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 11.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{b}.{\displaystyle y=2\sin \left({\displaystyle \frac{1}{2}\pi -x}\right)}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x+{\displaystyle \frac{1}{6}\pi })}\\ \\ & \text{d}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi +x}\right)}\\ \\ & \text{e}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi -2x}\right)}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}2\pi ,\text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (2x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kirinya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-2\sin (3x+45{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=-2\sin (3x-15{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=-2\sin (3x-45{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=2\sin (3x+15{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=2\sin (3x-45{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kanan}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{3}={120}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin 3(x-k)\\ & \text{dengan}-k\text{adalah}\\ & \mathbf{\text{besar geseran ke kanan}}{15}^{\circ }\\ & \text{Jadi},y=2\sin 3(x-15{)}^{\circ }=2\sin (3x-45{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-\cos (2x-30{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-60{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=\cos (2x+30{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=\sin (2x-80{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=\sin (2x+60{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}1\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=\cos 2(x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{15}^{\circ }\\ & \text{Jadi},y=\cos 2(x+15{)}^{\circ }=\sin (2x+30{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$

Lanjutan Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{positif terkecil yang memenuhi}\\ & \sin x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{adalah}....}\\ \\ & \text{a}.{\displaystyle {30}^{\circ }}\\ \\ & \text{b}.{\displaystyle {60}^{\circ }}\\ \\ & \text{c}.{\displaystyle {120}^{\circ }}\\ \\ & \text{d}.{\displaystyle {240}^{\circ }}\\ \\ & \text{e}.{\displaystyle {300}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin x& =-\frac{1}{2}\sqrt{3}\\ \text{Gun}& \text{akan rumus persamaan}\\ & \text{sederhana, yaitu}:\\ \sin x& =-\sin {60}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })\\ & =\sin {240}^{\circ }\\ x& ={240}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}\cos x={\displaystyle \frac{2\sqrt{5}}{5}\text{maka nilai}}\\ & \cot x\left({\displaystyle \frac{\pi }{2}-x}\right)\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{2}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}}\\ \\ & \text{c}.{\displaystyle \frac{1}{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{7}}\\ \\ & \text{e}.{\displaystyle \frac{1}{8}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos x& =\frac{2\sqrt{5}}{5},\text{maka}\\ {\sin }^{2}x& +{\cos }^{2}x=1\\ \sin x& =1-{\cos }^{2}x\\ & =\sqrt{1-{\cos }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{2\sqrt{5}}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{20}{25}}}=\sqrt{{\displaystyle \frac{5}{25}}}={\displaystyle \frac{\sqrt{5}}{5}}\\ \cot & \left({\displaystyle \frac{\pi }{2}-x}\right)=\tan x,\text{maka}\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}\\ & ={\displaystyle \frac{\sqrt{5}}{2\sqrt{5}}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Periode dari fungsi}f(x)=-2\cos 3x\text{adalah}....\\ \\ & \text{a}.{90}^{\circ }\\ \\ & \text{b}.{100}^{\circ }\\ \\ & \text{c}.{120}^{\circ }\\ \\ & \text{d}.{150}^{\circ }\\ \\ & \text{e}.{180}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Periode dari}:f(x)=-2\cos 3x\\ & \text{adalah}\\ & ={\displaystyle \frac{{360}^{\circ }}{3}}\\ & ={120}^{\circ }\\ & \\ & \text{Ingat bahwa}\\ & f(x)=a\cos bx,\text{maka periodenya}\\ & ={\displaystyle \frac{{360}^{\circ }}{b}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Perhatikanlah grafik berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Gambar di atas adalah grafik fungsi dari}\\ & \text{a}.f(x)=\cos 2x\\ \\ & \text{b}.f(x)=\cos 3x\\ \\ & \text{c}.f(x)=3\cos x\\ \\ & \text{d}.f(x)=3\cos 3x\\ \\ & \text{e}.f(x)={\displaystyle \frac{1}{3}\cos x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Gambar cukup jelas}\\ & \text{dengan periode}{360}^{\circ }\\ & \text{gambar dari grafik}f(x)=3\cos x\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}{\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}=....}\\ \\ & \text{a}.4\\ \\ & \text{b}.2-\sqrt{2}\\ \\ & \text{c}.\sqrt{2}-2\\ \\ & \text{d}.-4\\ \\ & \text{e}.-1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}}\\ & ={\displaystyle \frac{\sin {150}^{\circ }+\tan {180}^{\circ }+\cos {180}^{\circ }}{\sin {\displaystyle {90}^{\circ }+\cos {360}^{\circ }-3\cos {\displaystyle {60}^{\circ }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}+0+(-1)}}{1+1-3\left({\displaystyle \frac{1}{2}}\right)}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}}}{2-{\displaystyle \frac{3}{2}}}}\\ & =-{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =-1\end{array}\end{array}$

Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 1.& \text{Nilai}{75}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 75\times 1^{\circ }& =75\times {\displaystyle \frac{\pi }{180}radian}\\ {75}^{\circ }& ={\displaystyle \frac{5}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\tan \theta ={\displaystyle \frac{5}{12}\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{maka}\cos \theta \text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{5}{13}}\\ \\ & \text{b}.{\displaystyle \frac{12}{13}}\\ \\ & \text{c}.{\displaystyle \frac{13}{5}}\\ \\ & \text{d}.{\displaystyle \frac{13}{12}}\\ \\ & \text{e}.{\displaystyle \frac{12}{5}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah gambar segitiga berikut}\end{array}$.

$.\begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan \theta & ={\displaystyle \frac{5}{12},\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{lihat gambar di atas}\\ & \text{dengan dalil Pythagoras akan}\\ & \text{didapatkan sisimiringnya}=13\\ \text{jadi}& ,\text{nilai dari}\\ \cos \theta & ={\displaystyle \frac{12}{13}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Panjang BC adalah}....\\ & \text{a}.{\displaystyle 20\sin {36}^{\circ }}\\ & \text{b}.{\displaystyle 20\cos {36}^{\circ }}\\ & \text{c}.{\displaystyle 20\tan {36}^{\circ }}\\ & \text{d}.{\displaystyle 15}\\ & \text{e}.{\displaystyle 16}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan {36}^{\circ }& ={\displaystyle \frac{BC}{20}}\\ \iff & BC=20\tan {36}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\tan {300}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {300}^{\circ }& =\tan ({360}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}\tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{6}\sqrt{6}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{2}\sqrt{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\\ & =\tan {60}^{\circ }-\sin ({180}^{\circ }-{60}^{\circ })-\tan ({180}^{\circ }+{30}^{\circ })\\ & =\tan {60}^{\circ }-\sin {60}^{\circ }-\tan {30}^{\circ }\\ & =\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{3}\sqrt{3}}}\\ & =(1-{\displaystyle \frac{1}{2}-\frac{1}{3}})\sqrt{3}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}\end{array}$

Grafik Fungsi Trigonometri

Sebelumnya telah diketahui perbandingan trigonometri diberbagai kuadan dan sudut-sudut yang berelasi, selanjutnya dapat digambarkan garfik fungsinya, yaitu : $y=\sin x$, $y=\cos x$, dan $y=\tan x$.

A. Grafik Fungsi Sinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\sin b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

B. Grafik Fungsi Cosinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\cos b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

C. Grafik Fungsi Tangen

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\tan b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{180}^{\circ }}{b}\text{atau}{\displaystyle \frac{\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{nilai minimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& ={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{\displaystyle \frac{\pi }{6}})\text{atau boleh juga}}\\ & \text{dituliskan dengan bentuk}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{30}^{\circ })}\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|{\displaystyle \frac{4}{5}}\right|=\frac{4}{5}\\ \text{c}.& \text{Nilai minimumnya}:-\left|{\displaystyle \frac{4}{5}}\right|=-\frac{4}{5}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:{30}^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi}f(x)=2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & =2\cos 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =2\cos 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|2\right|=2\\ \text{c}.& \text{Nilai minimumnya}:-\left|2\right|=-2\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui fungsi}f(x)=\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & =\tan 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =\tan 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{180}^{\circ }}{2}}\right|={90}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\mathit{\text{tidak ada}}\\ \text{c}.& \text{Nilai minimumnya}:\mathit{\text{tidak ada}}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.