Giải thích các bước giải:

Câu 16:

a.Ta có:

$\sin\dfrac{7\pi}{12}=\sin(\dfrac13\pi+\dfrac14\pi)=\sin(\dfrac13\pi)\cos(\dfrac14\pi)+\cos(\dfrac13\pi)\sin(\dfrac14\pi)  =\dfrac{\sqrt{3}}{2}\cdot \:\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}\cdot \:\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}$

b.Ta có

$\cos\dfrac{5\pi}{12}+\cos\dfrac{\pi}{12}=2\cos\dfrac{\dfrac{5\pi}{12}+\dfrac{\pi}{12}}2\cdot \cos\dfrac{\dfrac{5\pi}{12}-\dfrac{\pi}{12}}2=2\cos \left(\dfrac{\pi }{4}\right)\cos \left(\dfrac{\pi }{6}\right)=2\cdot \:\dfrac{\sqrt{2}}{2}\cdot \:\dfrac{\sqrt{3}}{2}=\sqrt{\dfrac{3}{2}}$

c.Ta có:

$\sin\dfrac{\pi}{24}\sin\dfrac{5\pi}{24}=-\dfrac12(\cos(\dfrac{\pi}{24}+\dfrac{5\pi}{24})-\cos(\dfrac{5\pi}{24}-\dfrac{\pi}{24}))=-\dfrac12(\cos\dfrac{\pi}4-\cos\dfrac{\pi}6)=-\dfrac{\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}}{2}=-\dfrac{\sqrt{2}-\sqrt{3}}{4}$

Câu 17:

a.Ta có:

$\cos165^o=-\cos(180^o-165^o)=-\cos15^o=-\sqrt{\dfrac{\cos(2\cdot 15^o)-1}2}=-\sqrt{\dfrac{\cos30^o-1}2}=\dfrac{-\sqrt{6}-\sqrt{2}}{4}$

 b.Ta có:

$\tan\dfrac{7\pi}{12}=\tan(\dfrac13\pi+\dfrac14\pi)=\dfrac{\tan\dfrac13\pi+\tan\dfrac14\pi}{1-\tan\dfrac13\pi\tan\dfrac14\pi}=-2-\sqrt3$