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

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

$\text{CONTOH SOAL}$.

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

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

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

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

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

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

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