Question

Cho \(\pi< \alpha< \dfrac{3\pi}{2}\) và sin a = \(\dfrac{-5}{13}\) . Tính cosa , sin2a , cos2a , và sin\(\dfrac{a}{2}\)

Answer 1

Lời giải:

$\sin ^2a+\cos ^2a=1$

$\cos ^2a=1-\sin ^2a=1-(\frac{-5}{13})^2=\frac{144}{169}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\cos a< 0$

Do đó: $\cos a=-\sqrt{\frac{144}{169}}=\frac{-12}{13}$

$\sin 2a=2\sin a\cos a=2.\frac{-5}{13}.\frac{-12}{13}=\frac{120}{169}$

$\cos 2a=\cos ^2a-\sin ^2a=2\cos ^2a-1=2.\frac{144}{169}-1=\frac{119}{169}$

$\cos a=\cos ^2\frac{a}{2}-\sin ^2\frac{a}{2}$

$=1-2\sin ^2\frac{a}{2}$

$\Leftrightarrow \frac{-12}{13}=1-2\sin ^2\frac{a}{2}$

$\Rightarrow \sin ^2\frac{a}{2}=\frac{25}{26}$

Vì $\pi < a< \frac{3\pi}{2}$ nên $\sin \frac{a}{2}>0$

$\Rightarrow \sin \frac{a}{2}=\frac{5}{\sqrt{26}}$