Lanjutan 2 Materi Rumus-Rumus Trigonometri

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

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

$\begin{cases} \sin \left ( \alpha +\gamma \right ) & =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \sin \left ( \alpha -\gamma \right ) & =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \cos \left ( \alpha +\gamma \right ) & =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \end{cases}$.

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

$\begin{aligned} \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\sin \left ( \alpha +\beta \right )}{\cos \left ( \alpha +\beta \right )}\\ &=\displaystyle \frac{\left (\sin \alpha \cos \beta +\cos \alpha \sin \beta \right )\times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right ) }{\left (\cos \alpha \cos \beta -\sin \alpha \sin \beta \right ) \times \left ( \displaystyle \frac{1}{\cos \alpha \cos \beta } \right )}\\ &=\displaystyle \frac{\displaystyle \frac{\sin \alpha }{\cos \alpha }+\displaystyle \frac{\sin \beta }{\cos \beta }}{1-\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}\\ &=\displaystyle \color{red}\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\color{black}.\qquad \blacksquare \end{aligned}$.

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

$\begin{aligned} \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ &\qquad \textrm{dengan mengganti}\: \color{blue}\beta =-\beta \: \: \color{black}\textrm{maka,}\\ &=\displaystyle \frac{\tan \alpha +\tan \left ( -\beta \right )}{1-\tan \alpha \tan \left ( -\beta \right )}\\ \tan \left ( \alpha -\beta \right )&=\displaystyle \color{red}\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }\color{black}.\qquad \blacksquare \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Tunjukkan bahwa nilai}\: \: \tan 60^{\circ}=\color{red}\sqrt{3}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\tan 60^{\circ}&=\tan \left ( 30^{\circ}+30^{\circ} \right )\\ &=\displaystyle \frac{\tan 30^{\circ} +\tan 30^{\circ} }{1-\tan 30^{\circ} \tan 30^{\circ} }\\ &=\displaystyle \frac{2\tan 30^{\circ}}{1-\tan^{2} 30^{\circ}}\\ &=\displaystyle \frac{2\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )}{1-\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )^{2}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{1-\displaystyle \frac{1}{9}\sqrt{9}}\\ &=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{1-\displaystyle \frac{3}{9}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{\displaystyle \frac{6}{9}}=\displaystyle \frac{\displaystyle \frac{2}{3}\sqrt{3}}{\displaystyle \frac{2}{3}}\\ &=\displaystyle \color{red}\sqrt{3}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tunjukkan bahwa nilai}\\ & \tan 90^{\circ}=\color{red}\textrm{Tidak Terdefinisi}\\\\ &\color{blue}\textbf{Bukti}:\\ &\begin{aligned}\tan 90^{\circ}&=\tan \left ( 45^{\circ}+45^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ} +\tan 45^{\circ} }{1-\tan 45^{\circ} \tan 45^{\circ} }\\ &=\displaystyle \frac{2\tan 45^{\circ}}{1-\tan^{2} 45^{\circ}}\\ &=\displaystyle \frac{2\left ( 1 \right )}{1-\left ( 1 \right )^{2}}=\displaystyle \frac{\displaystyle 2}{1-1}\\ &=\displaystyle \frac{2}{0}\\ &=\color{red}\textrm{Tidak Terdefinisi}\qquad \color{black}\blacksquare \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukan nilai dari}\: \: \tan 75^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan 75^{\circ}&=\tan \left ( 30^{\circ}+45^{\circ} \right )\\ &=\displaystyle \frac{\tan 30^{\circ} +\tan 45^{\circ} }{1-\tan 30^{\circ} \tan 45^{\circ} }\\ &=\displaystyle \frac{\displaystyle \frac{1}{3}\sqrt{3}+1}{1-\left ( \displaystyle \frac{1}{3}\sqrt{3} \right )\left ( 1 \right )}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{3}\sqrt{3}}{1-\displaystyle \frac{1}{3}\sqrt{3}}=\displaystyle \frac{\displaystyle \frac{1}{3}\left (3+\sqrt{3} \right )}{\displaystyle \frac{1}{3}\left (3-\sqrt{3} \right )}\\ &=\displaystyle \frac{3+\sqrt{3} }{3-\sqrt{3} }\\ &=\displaystyle \frac{3+\sqrt{3} }{3-\sqrt{3} }\times \displaystyle \frac{3+\sqrt{3} }{3+\sqrt{3} }\\ &=\displaystyle \frac{3^{2}+3\sqrt{3}+3\sqrt{3}+\sqrt{9}}{3^{2}-\left ( \sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{12+6\sqrt{3}}{9-3}=\displaystyle \frac{12+6\sqrt{3}}{6}\\ &=\displaystyle \frac{6}{6}\left ( 2+\sqrt{3} \right )\\ &=\color{red}2+\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Tentukan nilai dari}\: \: \tan 105^{\circ}\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan 105^{\circ}&=\tan \left ( 45^{\circ}+60^{\circ} \right )\\ &=\displaystyle \frac{\tan 45^{\circ} +\tan 60^{\circ} }{1-\tan 45^{\circ} \tan 60^{\circ} }\\ &=\displaystyle \frac{1+\sqrt{3}+1}{1-1.\left ( \sqrt{3} \right )}\\ &=\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\\ &=\displaystyle \frac{1+\sqrt{3} }{1-\sqrt{3} }\\ &=\displaystyle \frac{1+\sqrt{3} }{1-\sqrt{3} }\times \displaystyle \frac{1+\sqrt{3} }{1+\sqrt{3} }\\ &=\displaystyle \frac{1^{2}+\sqrt{3}+\sqrt{3}+\left (\sqrt{3} \right )^{2}}{1^{2}-\left ( \sqrt{3} \right )^{2}}\\ &=\displaystyle \frac{4+2\sqrt{3}}{1-3}=\displaystyle \frac{4+2\sqrt{3}}{-2}\\ &=\color{red}-2-\sqrt{3} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Sederhanakan bentuk dari}\: \: \tan \left ( 270^{\circ}+A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan \left ( 270^{\circ}+A \right )&=\displaystyle \frac{\tan 270^{\circ}+\tan A}{1-\tan 270^{\circ}\tan A}\\ &=\displaystyle \frac{\textbf{TD}+\tan A}{1-\textbf{TD}\tan A}\\ &\quad \textrm{dengan}\: \: \textbf{TD}\: \textrm{adalah Tidak Terdefinisi}\\ &=\: \textrm{Bentuk yang harus dihindari}\\ \textrm{maka gunakan}&\: \textrm{bentuk berikut ini}\\ \tan \left ( 270^{\circ}+A \right )&=\displaystyle \frac{\sin \left ( 270^{\circ}+A \right )}{\cos \left ( 270^{\circ}+A \right )}\\ &=\displaystyle \frac{-\cos A}{\sin A}\\ &= \color{red}-\cot A\\\\ \color{purple}\textbf{Catatan}:&\: \textrm{lihatlah materi sebelumnya} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 6.&\textrm{Sederhanakan bentuk dari}\: \: \tan \left ( 270^{\circ}-A \right )\\\\ &\color{blue}\textbf{Jawab}:\\ &\begin{aligned}\tan \left ( 270^{\circ}-A \right )&=\displaystyle \frac{\sin \left ( 270^{\circ}-A \right )}{\cos \left ( 270^{\circ}-A \right )}\\ &=\displaystyle \frac{-\cos A}{-\sin A}\\ &= \color{red}\cot A \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Tunjukkan bahwa}\\ &\begin{array}{lll}\\ \textrm{a}.&\tan \left ( 2\alpha \right )=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha }\\ \textrm{b}.&\tan \alpha =\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan \left ( \displaystyle \frac{1}{2}\alpha \right )} \end{array}\\\\ &\color{blue}\textbf{Bukti}\\ &\begin{aligned}\textnormal{a.}\quad \tan \left ( \alpha +\beta \right )&=\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } ,&\textnormal{dan jika}\quad \beta =\alpha ,&&\textnormal{maka}\\ \tan \left ( \alpha +\alpha \right )&=\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha } \\ \tan 2\alpha &= \displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } \alpha\qquad \blacksquare \\ \textnormal{b.}\qquad\quad \tan 2\alpha &=\displaystyle \frac{2\tan \alpha }{1-\tan ^{2}\alpha } ,&\textnormal{dan jika}\quad \alpha =\displaystyle \frac{1}{2}\alpha ,&&\textnormal{maka}\\ \tan 2\left ( \displaystyle \frac{1}{2}\alpha \right )&=\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan ^{2}\left ( \displaystyle \frac{1}{2}\alpha \right )}\\ \tan \alpha &=\displaystyle \frac{2\tan \left ( \displaystyle \frac{1}{2}\alpha \right )}{1-\tan ^{2}\left ( \displaystyle \frac{1}{2}\alpha \right )}\qquad \blacksquare \end{aligned}\\ \end{array}$.