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:

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

$\text{CONTOH SOAL}$.

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

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

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

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