Question

Giúp mình câu này với

Biết \(\cos^2\)α - \(\sin^2\)α= \(\dfrac{1}{2}\) và \(\dfrac{3\pi}{2}\)≤α≤2\(\pi\) thì sin2α bằng

 

Answer 1

\(\dfrac{3\pi}{2}\le a\le2\pi\Rightarrow3\pi\le2a\le4\pi\)

\(\Rightarrow sin2a\le0\)

\(cos^2a-sin^2a=\dfrac{1}{2}\Leftrightarrow cos2a=\dfrac{1}{2}\)

\(\Rightarrow sin2a=-\sqrt{1-cos^22a}=-\dfrac{\sqrt{3}}{2}\)