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.

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:

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

$\text{CONTOH SOAL}$.

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

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

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

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

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

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

$\text{CONTOH SOAL}$.

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

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

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

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

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

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

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

Lanjutan 1 Materi Rumus-Rumus Trigonometri

A. 2  Rumus  $\cos (\alpha +\beta )$ dan $\cos (\alpha -\beta )$.

Dalam penentuan rumus $\cos (\alpha -\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Jika diurai gambar di atas adalah 

Mungkin gambarnya ada tang kurang jelas, mari kita perjelas lagi gambar di atas

Perhatikan bahwa

dan

Sehingga

$\begin{array}{rl}\left[ABC\right]& =\left[ACD\right]+\left[BCD\right]\\ {\displaystyle \frac{1}{2}ab\sin ({90}^0-\alpha +\beta )}& ={\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +{\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta }}\\ \sin ({90}^0-(\alpha -\beta ))& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos (\alpha -\beta )& =\cos \alpha \cos \beta +\sin \alpha \sin \beta .◼\end{array}$.

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

$\begin{array}{rl}& \cos (\alpha -(-\beta ))=\cos (\alpha +\beta )\\ & =\cos \alpha \cos (-\beta )+\sin \alpha \sin (-\beta )\\ & =\cos \alpha \cos \beta +\sin \alpha (-\sin \beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\cos {60}^{\circ }& =\cos ({30}^{\circ }+{30}^{\circ })\\ & =\cos {30}^{\circ }\cos {30}^{\circ }-\sin {30}^{\circ }\sin {30}^{\circ }\\ & ={\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }\\ & ={\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{1}{4}\sqrt{9}-\frac{1}{4}={\displaystyle \frac{3}{4}-\frac{1}{4}={\displaystyle \frac{2}{4}}}}\\ & ={\displaystyle \frac{1}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\cos {90}^{\circ }=0\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \cos {90}^{\circ }& =\cos ({60}^{\circ }+{30}^{\circ })\\ & =\cos {60}^{\circ }\cos {30}^{\circ }-\sin {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{3}-{\displaystyle \frac{1}{4}\sqrt{3}}}\\ & ={\displaystyle 0◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \cos {90}^{\circ }& =\cos ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \cos {90}^{\circ }& =\cos ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\cos {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {75}^{\circ }& =\cos ({30}^{\circ }+{45}^{\circ })\\ & =\cos {30}^{\circ }\cos {45}^{\circ }-\sin {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)-\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{6}-{\displaystyle \frac{1}{4}\sqrt{2}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{6}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\cos {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {105}^{\circ }& =\cos ({45}^{\circ }+{60}^{\circ })\\ & =\cos {45}^{\circ }\cos {60}^{\circ }-\sin {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}-{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}-\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }+A)& =\cos {270}^{\circ }\cos A-\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A-(-1)\sin A\\ & =0+\sin A\\ & =\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }-A)& =\cos {270}^{\circ }\cos A+\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A+(-1)\sin A\\ & =0-\sin A\\ & =-\sin A\end{array}\end{array}$.

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

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.

Rumus-Rumus Trigonometri

Sebelumnya kita buka arsip lama di blog ini, yaitu:

 Materi Lawas

A. Rumus Jumlah Dan selisih Sudut

A. 1  Rumus  $\sin (\alpha +\beta )$ dan $\sin (\alpha -\beta )$.

Dalam penentuan rumus $\sin (\alpha +\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Bukti:

$\begin{array}{rl}{\displaystyle \frac{AC}{\sin {90}^0}}& ={\displaystyle \frac{C{A}^{\prime }}{\sin \alpha }={\displaystyle \frac{A{A}^{\prime }}{\sin \mathrm{\angle }C}}}\\ A{A}^{\prime }& =AC.\sin \mathrm{\angle }C\\ & =AC.\sin ({90}^0-\alpha )\\ & =AC.\cos \alpha \\ & \text{dengan cara yang kurang lebih }\\ & \text{sama akan diperoleh juga}\\ A{A}^{\prime }& =AB.\cos \beta \\ & \text{selanjutnya kita tentukan luasnya, yaitu}\\ \left[ABC\right]& =\left[A{A}^{\prime }C\right]+\left[A{A}^{\prime }B\right]\\ {\displaystyle \frac{1}{2}.AB.AC.\sin (\alpha +\beta )}& ={\displaystyle \frac{1}{2}.AC.A{A}^{\prime }.\sin \alpha +{\displaystyle \frac{1}{2}.AB.A{A}^{\prime }.\sin \beta }}\\ \sin (\alpha +\beta )& ={\displaystyle \frac{AC.A{A}^{\prime }.\sin \alpha }{AB.AC}+{\displaystyle \frac{AB.A{A}^{\prime }.\sin \beta }{AB.AC}}}\\ & ={\displaystyle \frac{A{A}^{\prime }}{AB}.\sin \alpha +{\displaystyle \frac{A{A}^{\prime }}{AC}.\sin \beta }}\\ & ={\displaystyle \frac{(AB.\cos \beta )}{AB}.\sin \alpha +{\displaystyle \frac{(AC.\cos \alpha )}{AC}.\sin \beta }}\\ \sin (\alpha +\beta )& =\sin \alpha .\cos \beta +\cos \alpha .\sin \beta ◼\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\sin (\alpha -\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}& \sin (\alpha +(-\beta ))=\sin (\alpha -\beta )\\ & =\sin \alpha \cos (-\beta )+\cos \alpha \sin (-\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha (-\sin \beta )\\ & =\sin \alpha \cos \beta -\cos \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

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

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\sin {90}^{\circ }=1\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \sin {90}^{\circ }& =\sin ({60}^{\circ }+{30}^{\circ })\\ & =\sin {60}^{\circ }\cos {30}^{\circ }+\cos {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)+\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{9}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{3}{4}+{\displaystyle \frac{1}{4}}}\\ & ={\displaystyle \frac{4}{4}}\\ & ={\displaystyle 1◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \sin {90}^{\circ }& =\sin ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \sin {90}^{\circ }& =\sin ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\sin {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {75}^{\circ }& =\sin ({30}^{\circ }+{45}^{\circ })\\ & =\sin {30}^{\circ }\cos {45}^{\circ }+\cos {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\sin {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin {105}^{\circ }& =\sin ({45}^{\circ }+{60}^{\circ })\\ & =\sin {45}^{\circ }\cos {60}^{\circ }+\cos {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)+\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}+{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}+\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }+A)& =\sin {270}^{\circ }\cos A+\cos {270}^{\circ }\sin A\\ & =(-1)\cos A+\left(0\right)\sin A\\ & =-\cos A+0\\ & =-\cos A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\sin ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin ({270}^{\circ }-A)& =\sin {270}^{\circ }\cos A-\cos {270}^{\circ }\sin A\\ & =(-1)\cos A-\left(0\right)\sin A\\ & =-\cos A-0\\ & =-\cos A\end{array}\end{array}$.

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

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.

Lanjutan 3 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 16.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}\text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{75}^{\circ },{285}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{75}^{\circ },{343}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{75}^{\circ },{344}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{75}^{\circ },{345}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{75}^{\circ },{346}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{2}}\\ & \cos (x-30{)}^{\circ }=\cos {45}^{\circ }\\ & \iff x-{30}^{\circ }=\pm {45}^{\circ }+k{.360}^{\circ }\\ & \iff x={30}^{\circ }\pm {45}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={75}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{15}^{\circ }(tm)\\ & k=1\Rightarrow x_1={75}^{\circ }+{360}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{15}^{\circ }+{360}^{\circ }={345}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {75}^{\circ },{345}^{\circ }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos (x-30{)}^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}\text{untuk}{0}^{\circ }<x<{360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{{100}^{\circ },{330}^{\circ }\}}\\ \\ & \text{b}.{\displaystyle \{{30}^{\circ },{330}^{\circ }\}}\\ \\ & \text{c}.{\displaystyle \{{120}^{\circ },{300}^{\circ }\}}\\ \\ & \text{d}.{\displaystyle \{{60}^{\circ },{120}^{\circ }\}}\\ \\ & \text{e}.{\displaystyle \{{50}^{\circ },{300}^{\circ }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos x^{\circ }={\displaystyle \frac{1}{2}\sqrt{3}}\\ & \cos x^{\circ }=\cos {30}^{\circ }\\ & \iff x=\pm {30}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={30}^{\circ }(\text{mm})\\ & \text{atau}{x}_2=-{30}^{\circ }(tm)\\ & k=1\Rightarrow x_1={30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tm})\\ & \text{atau}{x}_2=-{30}^{\circ }+{360}^{\circ }={330}^{\circ }(mm)\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle {30}^{\circ },{330}^{\circ }}\right\}\end{array}\end{array}$

Lanjutan 2 Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 11.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{b}.{\displaystyle y=2\sin \left({\displaystyle \frac{1}{2}\pi -x}\right)}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x+{\displaystyle \frac{1}{6}\pi })}\\ \\ & \text{d}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi +x}\right)}\\ \\ & \text{e}.{\displaystyle y=-2\sin \left({\displaystyle \frac{1}{2}\pi -2x}\right)}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}2\pi ,\text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (2x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kirinya}}\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-2\sin (3x+45{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=-2\sin (3x-15{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=-2\sin (3x-45{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=2\sin (3x+15{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=2\sin (3x-45{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi sinus di geser ke}\mathbf{\text{kanan}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{3}={120}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin 3(x-k)\\ & \text{dengan}-k\text{adalah}\\ & \mathbf{\text{besar geseran ke kanan}}{15}^{\circ }\\ & \text{Jadi},y=2\sin 3(x-15{)}^{\circ }=2\sin (3x-45{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=-\cos (2x-30{)}^{\circ }}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-60{)}^{\circ }}\\ \\ & \text{c}.{\displaystyle y=\cos (2x+30{)}^{\circ }}\\ \\ & \text{d}.{\displaystyle y=\sin (2x-80{)}^{\circ }}\\ \\ & \text{e}.{\displaystyle y=\sin (2x+60{)}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}1\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ },\text{maka}}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=\cos 2(x+k)\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{15}^{\circ }\\ & \text{Jadi},y=\cos 2(x+15{)}^{\circ }=\sin (2x+30{)}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$

Lanjutan Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{positif terkecil yang memenuhi}\\ & \sin x=-{\displaystyle \frac{1}{2}\sqrt{3}\text{adalah}....}\\ \\ & \text{a}.{\displaystyle {30}^{\circ }}\\ \\ & \text{b}.{\displaystyle {60}^{\circ }}\\ \\ & \text{c}.{\displaystyle {120}^{\circ }}\\ \\ & \text{d}.{\displaystyle {240}^{\circ }}\\ \\ & \text{e}.{\displaystyle {300}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\sin x& =-\frac{1}{2}\sqrt{3}\\ \text{Gun}& \text{akan rumus persamaan}\\ & \text{sederhana, yaitu}:\\ \sin x& =-\sin {60}^{\circ }\\ & =\sin ({180}^{\circ }+{60}^{\circ })\\ & =\sin {240}^{\circ }\\ x& ={240}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika}\cos x={\displaystyle \frac{2\sqrt{5}}{5}\text{maka nilai}}\\ & \cot x\left({\displaystyle \frac{\pi }{2}-x}\right)\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{2}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}}\\ \\ & \text{c}.{\displaystyle \frac{1}{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{7}}\\ \\ & \text{e}.{\displaystyle \frac{1}{8}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos x& =\frac{2\sqrt{5}}{5},\text{maka}\\ {\sin }^{2}x& +{\cos }^{2}x=1\\ \sin x& =1-{\cos }^{2}x\\ & =\sqrt{1-{\cos }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{2\sqrt{5}}{5}}\right)}^2}\\ & =\sqrt{1-{\displaystyle \frac{20}{25}}}=\sqrt{{\displaystyle \frac{5}{25}}}={\displaystyle \frac{\sqrt{5}}{5}}\\ \cot & \left({\displaystyle \frac{\pi }{2}-x}\right)=\tan x,\text{maka}\\ \tan x& ={\displaystyle \frac{\sin x}{\cos x}}\\ & ={\displaystyle \frac{\sqrt{5}}{2\sqrt{5}}}\\ & ={\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Periode dari fungsi}f(x)=-2\cos 3x\text{adalah}....\\ \\ & \text{a}.{90}^{\circ }\\ \\ & \text{b}.{100}^{\circ }\\ \\ & \text{c}.{120}^{\circ }\\ \\ & \text{d}.{150}^{\circ }\\ \\ & \text{e}.{180}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Periode dari}:f(x)=-2\cos 3x\\ & \text{adalah}\\ & ={\displaystyle \frac{{360}^{\circ }}{3}}\\ & ={120}^{\circ }\\ & \\ & \text{Ingat bahwa}\\ & f(x)=a\cos bx,\text{maka periodenya}\\ & ={\displaystyle \frac{{360}^{\circ }}{b}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Perhatikanlah grafik berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Gambar di atas adalah grafik fungsi dari}\\ & \text{a}.f(x)=\cos 2x\\ \\ & \text{b}.f(x)=\cos 3x\\ \\ & \text{c}.f(x)=3\cos x\\ \\ & \text{d}.f(x)=3\cos 3x\\ \\ & \text{e}.f(x)={\displaystyle \frac{1}{3}\cos x}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Gambar cukup jelas}\\ & \text{dengan periode}{360}^{\circ }\\ & \text{gambar dari grafik}f(x)=3\cos x\end{array}$.

$\begin{array}{l}\\ 10.& \text{Nilai dari}{\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}=....}\\ \\ & \text{a}.4\\ \\ & \text{b}.2-\sqrt{2}\\ \\ & \text{c}.\sqrt{2}-2\\ \\ & \text{d}.-4\\ \\ & \text{e}.-1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {\displaystyle \frac{3}{4}\pi +\tan \pi +\cos \pi }}{\sin {\displaystyle \frac{1}{2}\pi +\cos 2\pi -3\cos {\displaystyle \frac{1}{3}\pi }}}}\\ & ={\displaystyle \frac{\sin {150}^{\circ }+\tan {180}^{\circ }+\cos {180}^{\circ }}{\sin {\displaystyle {90}^{\circ }+\cos {360}^{\circ }-3\cos {\displaystyle {60}^{\circ }}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}+0+(-1)}}{1+1-3\left({\displaystyle \frac{1}{2}}\right)}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{1}{2}}}{2-{\displaystyle \frac{3}{2}}}}\\ & =-{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =-1\end{array}\end{array}$

Soal Persiapan Penilaian Tengah Semester Gasal Kelas XI Pendalaman Matematika (Peminatan)

$\begin{array}{l}\\ 1.& \text{Nilai}{75}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 75\times 1^{\circ }& =75\times {\displaystyle \frac{\pi }{180}radian}\\ {75}^{\circ }& ={\displaystyle \frac{5}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Jika}\tan \theta ={\displaystyle \frac{5}{12}\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{maka}\cos \theta \text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{5}{13}}\\ \\ & \text{b}.{\displaystyle \frac{12}{13}}\\ \\ & \text{c}.{\displaystyle \frac{13}{5}}\\ \\ & \text{d}.{\displaystyle \frac{13}{12}}\\ \\ & \text{e}.{\displaystyle \frac{12}{5}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Perhatikanlah gambar segitiga berikut}\end{array}$.

$.\begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan \theta & ={\displaystyle \frac{5}{12},\text{untuk}{0}^{\circ }\le \theta \le {90}^{\circ }}\\ & \text{lihat gambar di atas}\\ & \text{dengan dalil Pythagoras akan}\\ & \text{didapatkan sisimiringnya}=13\\ \text{jadi}& ,\text{nilai dari}\\ \cos \theta & ={\displaystyle \frac{12}{13}}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Panjang BC adalah}....\\ & \text{a}.{\displaystyle 20\sin {36}^{\circ }}\\ & \text{b}.{\displaystyle 20\cos {36}^{\circ }}\\ & \text{c}.{\displaystyle 20\tan {36}^{\circ }}\\ & \text{d}.{\displaystyle 15}\\ & \text{e}.{\displaystyle 16}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ \tan {36}^{\circ }& ={\displaystyle \frac{BC}{20}}\\ \iff & BC=20\tan {36}^{\circ }\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai}\tan {300}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {300}^{\circ }& =\tan ({360}^{\circ }-{60}^{\circ })\\ & =-\tan {60}^{\circ }\\ & =-\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai}\tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \frac{1}{6}\sqrt{6}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{6}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{2}\sqrt{6}}\\ \\ & \text{d}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle \frac{1}{6}\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \tan {60}^{\circ }-\sin {120}^{\circ }-\tan {210}^{\circ }\\ & =\tan {60}^{\circ }-\sin ({180}^{\circ }-{60}^{\circ })-\tan ({180}^{\circ }+{30}^{\circ })\\ & =\tan {60}^{\circ }-\sin {60}^{\circ }-\tan {30}^{\circ }\\ & =\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{3}\sqrt{3}}}\\ & =(1-{\displaystyle \frac{1}{2}-\frac{1}{3}})\sqrt{3}\\ & ={\displaystyle \frac{1}{6}\sqrt{3}}\end{array}\end{array}$

Grafik Fungsi Trigonometri

Sebelumnya telah diketahui perbandingan trigonometri diberbagai kuadan dan sudut-sudut yang berelasi, selanjutnya dapat digambarkan garfik fungsinya, yaitu : $y=\sin x$, $y=\cos x$, dan $y=\tan x$.

A. Grafik Fungsi Sinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\sin b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

B. Grafik Fungsi Cosinus

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\cos b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{360}^{\circ }}{b}\text{atau}{\displaystyle \frac{2\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\left|a\right|\\ & \bullet \text{nilai minimum}:-\left|a\right|\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

C. Grafik Fungsi Tangen

Berikut ilustrasi grafik fungsi sinus untuk  $-\pi \le x\le \pi$.

$\begin{array}{rl}& \mathbf{\text{Bentuk umum}}\\ & f(x)=a\tan b(x+c)+d\\ & \bullet \text{periode}:{\displaystyle \frac{{180}^{\circ }}{b}\text{atau}{\displaystyle \frac{\pi }{\left|b\right|}}}\\ & \bullet \text{nilai maksimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{nilai minimum}:\mathit{\text{tidak ada}}\\ & \bullet \text{geseran grafik ke kiri}:c\\ & \bullet \text{geseran grafik ke kanan}:-c\\ & \bullet \text{geseran grafik ke atas}:d\\ & \bullet \text{geseran grafik ke bawah}:-d\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui fungsi}f(x)={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& ={\displaystyle \frac{4}{5}\sin (2x-{\displaystyle \frac{\pi }{3}})}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{\displaystyle \frac{\pi }{6}})\text{atau boleh juga}}\\ & \text{dituliskan dengan bentuk}\\ & ={\displaystyle \frac{4}{5}\sin 2(x-{30}^{\circ })}\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|{\displaystyle \frac{4}{5}}\right|=\frac{4}{5}\\ \text{c}.& \text{Nilai minimumnya}:-\left|{\displaystyle \frac{4}{5}}\right|=-\frac{4}{5}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:{30}^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Diketahui fungsi}f(x)=2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =2\cos (2x-{\displaystyle \frac{\pi }{4}})\\ & =2\cos 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =2\cos 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{360}^{\circ }}{2}}\right|={180}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\left|2\right|=2\\ \text{c}.& \text{Nilai minimumnya}:-\left|2\right|=-2\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Diketahui fungsi}f(x)=\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & \text{tentukanlah}\\ & \text{a}.\text{periode}\\ & \text{b}.\text{nilai maksimu}\\ & \text{c}.\text{nilai minimum}\\ & \text{d}.\text{arah geseran fungsinya}\\ & \text{e}.\text{gambarlah grafik fungsinya}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diket}& \text{ahui bahwa}\\ f(x)& =\tan (2x-{\displaystyle \frac{\pi }{4}})\\ & =\tan 2(x-{\displaystyle \frac{\pi }{8}})\text{atau boleh juga}\\ & \text{dituliskan dengan bentuk}\\ & =\tan 2(x-22,5^{\circ })\end{array}\\ & \begin{array}{rl}\text{a}.& \text{Periodenya}:\left|{\displaystyle \frac{{180}^{\circ }}{2}}\right|={90}^{\circ }\\ \text{b}.& \text{Nilai maksimumnya}:\mathit{\text{tidak ada}}\\ \text{c}.& \text{Nilai minimumnya}:\mathit{\text{tidak ada}}\\ \text{d}.& \text{Arah geserannya ke kanan sejauh}:22,5^{\circ }\\ \text{e}.& \text{Berikut gambar ilustrasinya}\end{array}\end{array}$.

Contoh Soal Numerasi Lanjutan 4 Persiapan Asesmen Nasional (AN)

$\text{CONTOH SOAL 14}$.

Perhatikanlah ilustrasi berikut

Jika sesorang ingin bepergian dari kota A ke kota C dengan melewati kota B dan jalur yang bisa dipilih tersedia sebagaimana pada gambar yaitu dari kota A ke kota B tersia 4 jalur berbeda dan dari kota B ke kota C tersedia 3 jalur berbeda, maka berapa jalur yang dapat ditempuh seseorang jika ia bermula dari kota A ke kota C dengan harus melewati kota B

$\text{SOLUSI SOAL 14}$.

Jalur yang dapat ditempuh adalah sebanyak 4 x 3 = 12 jalur dari kota A ke kota C dengan harus melewati kota B sebagaimana gambar pada soal di atas.

$\text{CONTOH SOAL 15}$.

Jika seseorang tadi berangkat dari kota A ke kota C dengan harus melewati kota C kemudian pulangnya dari C ke A dengan melewati kota B juga, berapa banyak jalur yang dapat dipilih jika antara berangkat dan pulang menggunakan jalur yang berbeda?

$\text{SOLUSI SOAL 15}$.

Dengan kondisi seperti pada soal karena saat berangkat semua jalur dapat dipilih semua dari daftar pilihan yang mungkin atau yang ada dan jika ketika pulang disyaratkan harus dengan memilih jalur yang berbeda dari saat berangkat, maka ketika pulang dari kota C akan mau ke B, maka jalur akan berkurang satu (ingat saat berangkat salah satu jalur ini sudah dipilih). Demikian pula saat sampai di B kemudian ingin menuju ke A, maka ia hanya punya 3 pilihan (karena saat berangkat ia punya 4 pilihan dan saat itu ia sudah mengambil satu jalur dari 4 jaur yang ada). Sehingga total jalur yang dapat dipilih ada sebanyak : 4 x 3 x (3-1) x (4-1) = 72 jalur yang berbeda.

Contoh Soal Numerasi Lanjutan 3 Persiapan Asesmen Nasional (AN)

 $\text{CONTOH SOAL 13}$.

Diberikan dua kertas karton A dan B yang sama ukurannya diatur dengan posisi berbeda berikut

Jika lebarnya (bagian seperti tinggi) baik kertas A maupun kertas B bagian tepinya direkatkan jadilah ia sebuah tabung yang tentunya tinggi tabung dari kertas A dan B akan menyesuaikan lebar kertasnya. Tentukanlah volume terbesar dari kedua tabung tersebut dan berilah alasannya

$\text{SOLUSI SOAL 13}$.

Perhatikanlah ilustrasi dari dua kertas yang sama ukurannya di atas yang dikondisikan posisinya berbeda

Lingkaran  di bawah masing-masing hanya menunjukkan bahwa setelah masing-masing kertas bagian lebar tepinya saling direkatkan akan berupa lingkaran karena hasilnya adalah berupa tabung (tentunya kedua tabungnya tanpa alas dan atap)

$\begin{array}{|lll|}\hline \text{Bagian}& \mathbf{\text{Lingkaran A}}& \mathbf{\text{Lingkaran B}}\\ \begin{array}{rl}& \text{keliling}\\ & \text{alas}\end{array}& 35& 14\\ \text{Jari-Jari}& \begin{array}{rl}2\pi r_{{}_A}& =35\\ r_{{}_A}& ={\displaystyle \frac{35}{2\pi }}\end{array}& \begin{array}{rl}2\pi r_{{}_B}& =14\\ r_{{}_B}& ={\displaystyle \frac{14}{2\pi }}\end{array}\\ \text{Luas}& \begin{array}{rl}\mathbf{\text{L}}& =\pi r_A^2\\ & =\pi {\left({\displaystyle \frac{35}{2\pi }}\right)}^2\\ & ={\displaystyle \frac{1225}{4\pi }}\end{array}& \begin{array}{rl}\mathbf{\text{L}}& =\pi r_B^2\\ & =\pi {\left({\displaystyle \frac{14}{2\pi }}\right)}^2\\ & ={\displaystyle \frac{196}{4\pi }}\end{array}\\ \hline\end{array}$.

Jelas bahwa pengaruh utama terletak pada panjang jari-jarinya. Karena dalam penghitungan volume jari-jarinya harus dikuadratkan, maka hasil yang didapatkan akan sangat berpengaruh. Dan karena jari-jari didasarkan berasal dari keliling lingkaran (panjang kertas), maka kertas dengan keliling alas terbesar akan memiliki volume yang terbesar pula.

Contoh Soal Numerasi Lanjutan 2 Persiapan Asesmen Nasional (AN)

Materi Pendukung

A. Faktor Pembilang

Misalkan kita ingin mendaftar bilangan keliapatan genap positif  $m$ yang kurang dari atau sama dengan $n$. 

Jika $n$ kelipatan dari $m$, maka akan dapat dituliskan $n=k.m$. Karena $n=k.m$ selanjutnya  $k={\displaystyle \frac{n}{m}}$ dan oleh karenanya dapat dituliskan pula $m,2m,\cdots ,km=n$.

Akan tetapi, jika $n$ bukan merupakan kelipatan dari $m$, maka akan terdapat suatu bilangan bulat $k$ dengan  $km<n<(k+1)m$, dengan  $k$ adalah bilangan bulat terbesar yang kurang dari  ${\displaystyle \frac{n}{m}}$.

B. Fungsi Bilangan Dasar/Fungsi Tangga

Untuk suatu bilangan asli  $x$, fungsi dasar  $x$ dinotasikan dengan  $\lfloor x\rfloor$ adalah bilangan bulat terbesar yang kurang dari atau sama dengan  $x$.

Contoh:

$\lfloor 1,7\rfloor =1,\lfloor \sqrt{5}\rfloor =2,\text{dan}\lfloor 4\rfloor =4$.

Selanjutnya, untuk menuliskan banyaknya bilangan bulat positif  $m$ yang kurang dari atau sama dengan bilangan bulat $n$  adalah  $\lfloor {\displaystyle \frac{n}{m}}\rfloor$.

Contoh:

Untuk menuliskan banyaknya bilangan kelipatan 3 yang yang terletak di antara bilangan 5 dan 10 dapat ditentukan dengan 

$\begin{array}{rl}& \lfloor {\displaystyle \frac{10}{3}}\rfloor -\lfloor {\displaystyle \frac{5}{3}}\rfloor =3-1=2\\ & \\ & \text{bukan dituliskan dengan}\\ & \lfloor {\displaystyle \frac{10-5}{3}}\rfloor =\lfloor {\displaystyle \frac{5}{3}}\rfloor =1\end{array}$.

Prinsip Inklusi Eksklusi

Hal yang berkaitan dengan jumlah hitungan dan tidaknya.

Jika diberikan $N$ objek, sebagai misal

• $N(\alpha )$  banyaknya objek dengan sifat  $\alpha$
• $N(\beta )$  banyaknya objek dengan sifat  $\beta$
• $N(\gamma )$  banyaknya objek dengan sifat  $\gamma$
• dan seterusnya

dan misal

• $N(\alpha ,\beta )$  banyaknya objek dengan sifat  $\alpha$  dan  $\beta$
• $N(\alpha ,\gamma )$  banyaknya objek dengan sifat  $\alpha$  dan  $\gamma$
• $N(\beta ,\gamma )$  banyaknya objek dengan sifat  $\beta$  dan  $\gamma$
• dan seterusnya

serta misalkan juga 

• $N(\alpha ,\beta ,\gamma )$  banyaknya objek dengan sifat  $\alpha$,  $\beta$ serta $\gamma$

Maka gabungan objek $N$ dengan sifat $\alpha$,  $\beta$ serta $\gamma$  adalah:

$\begin{array}{rl}& \left|N(\alpha )\cup N(\beta )\cup N(\gamma )\cup \cdots \right|\\ & =\left|N(\alpha )+N(\beta )+N(\gamma )+\cdots \right|\\ & -\left|N(\alpha ,\beta )+N(\alpha ,\gamma )+N(\beta ,\gamma )+\cdots \right|\\ & +\left|N(\alpha ,\beta ,\gamma )+\cdots \right|\\ & -\cdots \end{array}$.

$\text{CONTOH SOAL 11}$.

Perhatikanlah tabel berikut

Ada berapa banyak bilangan 1 - 500 yang habis dibagi 2 atau 3?

$\text{SOLUSI SOAL 11}$.

$\begin{array}{rl}S& =\text{himpunan bilangan antara}1-500\\ \text{m}& \text{isalkan}\\ A& =\text{himpunan bilangan habis dibagi 2}\\ B& =\text{himpunan bilangan habis dibagi 3}\\ A& \cup B=\text{himpunan bilangan habis dibagi 2 atau 3}\\ A& \cap B=\text{himpunan bilangan habis dibagi 2 dan 3}\\ \text{m}& \text{aka}\\ & |A\cup B|\\ & =\left|A\right|+\left|B\right|-|A\cap B|\\ & =\lfloor {\displaystyle \frac{500}{2}}\rfloor +\lfloor {\displaystyle \frac{500}{3}}\rfloor -\lfloor {\displaystyle \frac{500}{6}}\rfloor \\ & =250+166-83\\ & =333\end{array}$.

$\text{CONTOH SOAL 12}$.

Perhatikan pula tabel berikut

Ada berapa banyak pula bilangan 1 - 500 yang habis dibagi 2 atau 3 atau 5?

$\text{SOLUSI SOAL 12}$.

$\begin{array}{rl}S& =\text{himpunan bilangan antara}1-500\\ \text{m}& \text{isalkan}\\ A& =\text{himpunan bilangan habis dibagi 2}\\ B& =\text{himpunan bilangan habis dibagi 3}\\ C& =\text{himpunan bilangan habis dibagi 5}\\ A& \cup B\cup C=\text{himpunan bilangan habis dibagi 2, 3, atau 5}\\ A& \cap B=\text{himpunan bilangan habis dibagi 2 dan 3}\\ A& \cap C=\text{himpunan bilangan habis dibagi 2 dan 5}\\ B& \cap C=\text{himpunan bilangan habis dibagi 3 dan 5}\\ A& \cap B\cap C=\text{himpunan bilangan habis dibagi 2 dan 3 serta 5}\\ \text{m}& \text{aka}\\ & |A\cup B\cup C|\\ & =\left|A\right|+\left|B\right|+\left|C\right|-|A\cap B|-|A\cap C|\\ & -|B\cap C|+|A\cap B\cap C|\\ & =\lfloor {\displaystyle \frac{500}{2}}\rfloor +\lfloor {\displaystyle \frac{500}{3}}\rfloor +\lfloor {\displaystyle \frac{500}{5}}\rfloor -\lfloor {\displaystyle \frac{500}{6}}\rfloor -\lfloor {\displaystyle \frac{500}{10}}\rfloor -\lfloor {\displaystyle \frac{500}{15}}\rfloor +\lfloor {\displaystyle \frac{500}{30}}\rfloor \\ & =250+166+100-83-50-33+16\\ & =366\end{array}$.