Sina a / 2cos a = 2/3 tính sin a , cos a

Đáp án:

$\left[ \begin{array}{l} \cos a = - \frac{3}{5};\sin a = - \frac{4}{5}\\ \cos a = \frac{3}{5};\sin a = \frac{4}{5} \end{array} \right.$

Giải thích các bước giải: Áp dụng: ${\sin ^2}x + {\cos ^2}x = 1$ Ta có: $\begin{array}{l} \frac{{\sin a}}{{2\cos a}} = \frac{2}{3} \Leftrightarrow \frac{{\sin a}}{{\cos a}} = \frac{4}{3}\\ \Leftrightarrow \sin a = \frac{4}{3}\cos a \end{array}$ Do đó $\begin{array}{l} {\sin ^2}a + {\cos ^2}a = 1 \Leftrightarrow {\left( {\frac{4}{3}\cos a} \right)^2} + {\cos ^2}a = 1\\ \Leftrightarrow \frac{{25}}{9}{\cos ^2}a = 1\\ \Leftrightarrow {\cos ^2}a = \frac{9}{{25}}\\ \Leftrightarrow \left[ \begin{array}{l} \cos a = - \frac{3}{5};\sin a = - \frac{4}{5}\\ \cos a = \frac{3}{5};\sin a = \frac{4}{5} \end{array} \right.\\ \end{array}$

$\dfrac{\sin a}{2\cos a}=\dfrac{2}{3}$

$\Leftrightarrow \dfrac{\tan a}{2}=\dfrac{2}{3}$

$\Leftrightarrow \tan a=\dfrac{4}{3}$

$\dfrac{1}{\cos^2a}=1+\tan^2a$

$\to \cos a=\dfrac{3}{5}$

$\sin a=\sqrt{1-\cos^2a}=\dfrac{4}{5}$