B. 1 Rumus Sudut Rangkap
Pada materi sebelumnya sudah dibahas tentang hal ini yaitu,
$\begin{array}{ll}\\ \bullet &\sin 2\alpha =2\sin \alpha \cos \alpha \\ \bullet &\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha \\ \bullet &\tan 2\alpha =\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \end{array}$.
B. 2 Rumus Sudut Rangkap Tiga
Berikut untuk rumus sudut rangkap tiga:
$\begin{aligned}\bullet \: \sin 3\alpha &=3\sin \alpha -4\sin ^{3}\alpha \\ \bullet \: \cos 3\alpha &=4\cos ^{3}\alpha -3\cos \alpha \\ \bullet \: \tan 3\alpha &=\displaystyle \frac{3\tan x-\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \end{aligned}$
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:
$\begin{array}{ll}\\ \bullet &\sin \alpha =2\sin \displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha \\ \bullet &\cos \alpha =\cos ^{2}\displaystyle \frac{1}{2}\alpha -\sin ^{2}\displaystyle \frac{1}{2}\alpha \\ \bullet &\tan \alpha =\displaystyle \frac{2\tan \displaystyle \frac{1}{2}\alpha }{1-\tan ^{2}\displaystyle \frac{1}{2}\alpha } \end{array}$.
$\LARGE\colorbox{yellow}{CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \sin A=\displaystyle \frac{4}{5},\\ & \textrm{untuk}\: A\: \textrm{sudut tumpul. Tentukanlah}\\ &\textrm{a}.\quad \sin 2A\\ &\textrm{b}.\quad \cos 2A\\ &\textrm{c}.\quad \tan 2A\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Dengan menggunakan identitas}\\ &\sin ^{2}\alpha +\cos ^{2}\alpha =1,\: \: \textrm{dapat diperoleh}\\ &\textrm{nilai}\: \: \cos A,\: \: \textrm{yaitu}:\\ &\cos ^{2}\alpha =1-\sin ^{2}A=1-\left ( \displaystyle \frac{4}{5} \right )^{2}\\ &\qquad =1-\displaystyle \frac{16}{25}=\frac{9}{25}\Rightarrow \cos A=\pm \displaystyle \frac{3}{5}\\ &\textrm{ambil}\: \: \cos A=-\displaystyle \frac{3}{5},\: \: \textrm{karena di kuadran II}\\ &\textrm{ingat sudut tumpul}=\textrm{sudut di kuadran II}\\ &\begin{aligned}\textrm{a}.\: \: \sin 2A&=2\sin A\cos A\\ &=2\left ( \displaystyle \frac{4}{5} \right )\left ( -\displaystyle \frac{3}{5} \right )\\ &=-\displaystyle \frac{24}{25}\\ \textrm{b}.\: \: \cos 2A&=\cos ^{2}A-\sin ^{2}A\\ &=\left ( -\displaystyle \frac{3}{5} \right )^{2}-\left ( \displaystyle \frac{4}{5} \right )^{2}\\ &=\displaystyle \frac{9}{25}-\frac{16}{25}=-\frac{7}{25}\\ \textrm{c}.\: \: \tan 2A&=\displaystyle \frac{\sin 2A}{\cos 2A}\\ &=\displaystyle \frac{-\displaystyle \frac{24}{25}}{-\displaystyle \frac{7}{25}}=\displaystyle \frac{24}{7} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 2.& \textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \sin 15^{\circ}\\ &\textrm{b}.\quad \cos 15^{\circ}\\ &\textrm{c}.\quad \tan 15^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\textrm{Banyak cara yang bisa digunakan}.\\ &\textrm{berikut salah satu cara itu, yaitu}\\ &\begin{aligned}\textnormal{a.}\quad\quad \cos \left (2\alpha \right )&=1-2\sin ^{2}\alpha, &\textnormal{pilih untuk}\quad \alpha =15^{0} \\ \cos \left (2\times 15^{0} \right )&=1-2\sin^{2} 15^{0}\\ \cos 30^{0}&=1-2\sin ^{2}15^{0}\\ \displaystyle \frac{1}{2}\sqrt{3}&=1-2\sin ^{2}15^{0}\\ \sqrt{3}&=2-4\sin ^{2}15^{0}\\ \sin ^{2}15^{0}&=\displaystyle \frac{2-\sqrt{3}}{4}\\ \sin 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}\\ \textnormal{b.}\quad\quad \sin ^{2}\alpha +&\cos ^{2}\alpha =1,&\textnormal{rumus identitas}\\ \sin^{2}15^{0}+&\cos ^{2}15^{0}=1\\ \cos ^{2}15^{0}&=1-\sin ^{2}15^{0}\\ &=\displaystyle 1-\left ( \frac{2-\sqrt{3}}{4} \right ),&\textnormal{lihat jawaban poin a)}\\ &=\displaystyle \frac{2+\sqrt{3}}{4}\\ \cos 15^{0}&=\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}\\ \textnormal{c.}\qquad\quad \tan \alpha &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ \tan 15^{0}&=\displaystyle \frac{\sin 15^{0}}{\cos 15^{0}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}{\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}\\ &=\sqrt{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}\times \sqrt{\frac{2-\sqrt{3}}{2-\sqrt{3}}},&\textnormal{sekawan dari penyebut}\\ &=\sqrt{\displaystyle \frac{4-2.2\sqrt{3}+3}{4-3}}\\ &=\sqrt{7-4\sqrt{3}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tunjukkan bahwa nilai}\\ & \displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }=\color{red}\tan \alpha \\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }&=\displaystyle \frac{1-\left ( \cos ^{2}\alpha -\sin ^{2}\alpha \right ) }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1-\left ( \left ( 1-\sin ^{2}\alpha \right )-\sin ^{2}\alpha \right )}{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{1-1+\sin ^{2}\alpha +\sin ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{2\sin ^{2}\alpha }{2\sin \alpha \cos \alpha }\\ &=\displaystyle \frac{\sin \alpha }{\cos \alpha }\\ &=\displaystyle \color{red}\tan \alpha \qquad \color{black}\blacksquare \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Tunjukkan bahwa nilai}\\ &\textrm{a}.\quad \cos 3\alpha =4\cos ^{3}\alpha -3\cos \alpha \\ &\textrm{b}.\quad \tan 3\alpha =\displaystyle \frac{3\tan x-\tan ^{3}\alpha }{1-3\tan ^{2}\alpha } \\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad \cos 3\alpha &=\cos \left ( 2\alpha +\alpha \right )\\ &=\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ &=\left ( 2\cos ^{2}\alpha -1 \right )\cos \alpha -\left ( 2\sin \alpha \cos \alpha \right )\sin \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\sin ^{2}\alpha \cos \alpha \\ &=2\cos ^{3}\alpha -\cos \alpha -2\left ( 1-\cos ^{2}\alpha \right ) \cos \alpha\\ &=2\cos ^{3}\alpha -\cos \alpha -2\cos \alpha +2\cos ^{2}\alpha\\ &=4\cos ^{3}\alpha -3\cos \alpha \qquad \blacksquare\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \tan 3\alpha &=\tan \left ( 2\alpha +\alpha \right )\\ &=\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }\\ &=\displaystyle \frac{\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right )+\tan \alpha }{1-\left ( \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \right ).\tan \alpha }\\ &=\displaystyle \frac{\displaystyle \frac{2\tan \alpha +\tan \alpha -\tan ^{3}\alpha }{1-\tan ^{2}\alpha }}{\displaystyle \frac{\left ( 1-\tan ^{2}\alpha \right )-2\tan ^{2}\alpha }{1-\tan ^{2}\alpha }}\\ &=\displaystyle \frac{3\tan \alpha -\tan ^{3}\alpha }{1-3\tan ^{2}\alpha }\qquad \blacksquare \end{aligned} \\\hline \end{array} \end{array}$
Tidak ada komentar:
Posting Komentar
Informasi