Giải thích các bước giải:
Cho $\cos\alpha=\dfrac5{13} , (0<\alpha<\dfrac{\pi}{2})$
Tính:
a.$\cos(10\pi+\alpha), \cot (\dfrac{17\pi}{2}+\alpha)$
Ta có:
$0<\alpha<\dfrac{\pi}{2}\to \sin\alpha>0$
Mà $\sin^2\alpha=1-\cos^2\alpha=\dfrac{144}{169}\to \sin\alpha=\dfrac{12}{13}$
$\to\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{12}{5}$
Ta có:
$\cos(10\pi+\alpha)=\cos(\alpha)=\dfrac5{13}$
$\cot(\dfrac{17\pi}{2}+\alpha)$
$=\cot(8\pi+\dfrac{\pi}{2}+\alpha)$
$=\cot(\dfrac{\pi}{2}+\alpha)$
$=\tan(-\alpha)$
$=-\tan(\alpha)$
$=-\dfrac{12}5$
b.Ta có:
$\sin2\alpha=2\sin\alpha\cos\alpha=\dfrac{60}{169}$
$\cos2\alpha=2\cos^2\alpha-1=-\dfrac{119}{169}$
