SBMPTN Zone : Jumlah dan Selisih Trigonometri

Berikut ini adalah kumpulan soal mengenai jumlah dan selisih trigonometri tipe SBMPTN. Jika ingin bertanya soal, silahkan gabung ke grup Facebook atau Telegram.

Tipe:

Standar SBMPTN HOTS

No. 1

Jika \cos\alpha\cdot\cos \beta = \dfrac38 dan {\cos \left(\alpha + \beta\right) = \dfrac12} maka {\tan \left(\alpha-\beta\right) =}

1. \dfrac14
2. \dfrac12\sqrt3
3. \dfrac13\sqrt5

4. \sqrt{15}
5. \dfrac12\sqrt{15}

SKMP Juli 2020

$\begin{array}{rll}\cos \alpha \cdot \cos \beta & ={\displaystyle \frac{3}{8}}& \times 2\\ 2\cos \alpha \cdot \cos \beta & ={\displaystyle \frac{3}{4}}\\ \cos (\alpha +\beta )+\cos (\alpha -\beta )& ={\displaystyle \frac{3}{4}}\\ {\displaystyle \frac{1}{2}}+\cos (\alpha -\beta )& ={\displaystyle \frac{3}{4}}\\ \cos (\alpha -\beta )& ={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{2}}\\ \cos (\alpha -\beta )& ={\displaystyle \frac{1}{4}}\end{array}$

\sqrt{4^2-1^2}=\sqrt{!6-1}=\sqrt{!5}

$\begin{array}{rl}\tan (\alpha -\beta )& ={\displaystyle \frac{\sqrt{15}}{1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{15}}}}}\end{array}$

Jadi, {\tan \left(\alpha-\beta\right) =\sqrt{15}}.
JAWAB: D