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.