Trigonometri: Persiapan Masuk Perguruan Tinggi Negeri

Soal 1. SBMPTN 2017
Jika $x$ memenuhi $-2\csc{x}+2\cot{x}+3 \sin{x}=0$ untuk $0{^{\circ}} < x < \pi$, maka $\cos{x}= ....$
A. $-\frac{2}{3}$
B. $-\frac{1}{2}$
C. $-\frac{1}{3}$
D. $\frac{1}{2}$
E. $\frac{2}{3}$
Pembahasan: C
$-2\csc{x}+2\cot{x}+3 \sin{x}=0$
$-\frac{2}{\sin{x}}+\frac{2 \cos{x}}{\sin{x}}+3 \sin{x}=0$ (kalikan kedua ruas dengan $\sin{x}$)
$-2 + 2 \cos{x} + 3 \sin^{2}{x}=0$
$-2 + 2 \cos{x} + 3 (1-\cos^{2}{x})=0$
$-2 + 2 \cos{x} + 3- 3\cos^{2}{x}=0$
$-3\cos^{2}{x} + 2\cos{x}+1=0$
$3\cos^{2}{x} - 2\cos{x}-1=0$
$(3\cos{x}+1)(\cos{x}-1)=$
$\cos{x}=-\frac{1}{3}$ atau $\cos{x}=1$ (TM, karena $0{^{\circ}} < x < \pi$).

Soal 2.
Jika $\sin{\alpha} + \sin{\beta} = \sqrt{2A}$ dan $\cos{\alpha} + \cos{\beta} = \sqrt{2B}$, maka nilai dari $\cos(\alpha - \beta)=....$
A. $A+B-1$
B. $\frac{A+B-1}{2}$
C. $A+B-2$
D. $\frac{A+B-2}{2}$
E. $\frac{A+B-2}{4}$
Pembahasan: A
$\sin{\alpha} + \sin{\beta} = \sqrt{2A}$
$\sin^2{\alpha} +2\sin{\alpha}\sin{\beta} + \sin^2 {\beta} = 2A \ .............  pers. (1)$

$\cos{\alpha} + \cos{\beta} = \sqrt{2B}$
$\cos^2{\alpha} +2\cos{\alpha}\cos{\beta} + \cos^2 {\beta} = 2B \ .......... pers. (2)$

Jumlahkan persamaan (1) dan (2)
$\sin^2{\alpha}+\cos^2{\alpha}+2\sin{\alpha}\sin{\beta}+2\cos{\alpha}\cos{\beta}+\sin^2 {\beta}+\cos^2 {\beta}=2A+2B$
$1+2\sin{\alpha}\sin{\beta}+2\cos{\alpha}\cos{\beta}+1=2A+2B$
$2\sin{\alpha}\sin{\beta}+2\cos{\alpha}\cos{\beta}+2=2A+2B$
$2\sin{\alpha}\sin{\beta}+2\cos{\alpha}\cos{\beta}=2A+2B-2$
$\sin{\alpha}\sin{\beta}+\cos{\alpha}\cos{\beta}=A+B-1$
$\cos{(\alpha - \beta)}=A+B-1$

Baca Juga Vektor: Soal-Soal SBMPTN

Soal 3.
Diketahui $\sin{\alpha} \cos{\alpha} = \frac{8}{25}$. Nilai $\frac{1}{\sin{\alpha}} - \frac{1}{\cos{\alpha}} = ....$
A. \frac{3}{25}
B. \frac{9}{25}
C. \frac{5}{8}
D. \frac{3}{5}
E. \frac{15}{8}
Pembahasan: E
$\frac{1}{\sin{\alpha}} - \frac{1}{\cos{\alpha}} = \frac{\cos{\alpha}-\sin{\alpha}}{\sin{\alpha} \cos{\alpha}}$

Misalkan,
$\cos{\alpha}-\sin{\alpha}=a$
$(\cos{\alpha}-\sin{\alpha})^2=a^2$
$\cos^2 {\alpha}-2\cos{\alpha} \sin{\alpha}+\sin^2 {\alpha}=a^2$
$\sin^2 {\alpha}+\cos^2 {\alpha}-2\cos{\alpha} \sin{\alpha}=a^2$
$1-2(\frac{8}{25})=a^2$
$1-\frac{16}{25}=a^2$
$\frac{9}{25}=a^2$
$a=\frac{3}{5}$

$\frac{1}{\sin{\alpha}} - \frac{1}{\cos{\alpha}} = \frac{\cos{\alpha}-\sin{\alpha}}{\sin{\alpha} \cos{\alpha}}$
$\frac{1}{\sin{\alpha}} - \frac{1}{\cos{\alpha}} = \frac{\frac{3}{5}}{\frac{8}{25}}$
$\frac{1}{\sin{\alpha}} - \frac{1}{\cos{\alpha}} = \frac{15}{8}$

Persamaan Kuadrat
Logaritma
Polinomial (Suku Banyak)
Vektor