Tìm tan

D

demon311

$\cos{x}=\dfrac{-4}{5} $

$\pi < x < \dfrac{3\pi}{2}$ nên $x \in $ cung phần tư thứ 3
=> $\sin{x}, \cos{x}<0$
=> $\sin{x}=-\dfrac{3}{5} $ => $\tan{x}=\dfrac{3}{4}$
$\tan{(x+\dfrac{2015\pi}{10})}=\tan{(x+\dfrac{403 \pi}{2})}= \tan{(x+\dfrac{3\pi}{2}+100.2\pi)} \\
=\tan{(x+\dfrac{3\pi}{2})}=\dfrac{\sin{(x+\dfrac{3\pi}{2})}}{\cos{(x+\dfrac{3\pi}{2})}} =\dfrac{\sin{x}\cos{(\dfrac{3 \pi}{2})}+\cos{x}\sin{(\dfrac{3 \pi}{2})}}{\cos{x}\cos{(\dfrac{3\pi}{2})}+\sin{x} \sin{(\dfrac{3\pi}{2})}} \\
=\dfrac{-\dfrac{4}{5}.(-1)}{-\dfrac{3}{5}.(-1)}=\dfrac{4}{3}$
Hoặc:
$\tan{(x+\dfrac{3\pi}{2})}=\cot{(\pi-x)}=\cot{(-x)}=\cot{x}=\dfrac{1}{\tan{x}}=\dfrac{4}{3}$
 
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