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Question: If \(\sin\alpha = \frac{- 3}{5},\) where \(\pi < \alpha < \frac{3\pi}{2},\) then \(\cos\frac{1}{2}\a...

If sin⁡α=−35,\sin\alpha = \frac{- 3}{5}, where π<α<3π2,\pi < \alpha < \frac{3\pi}{2}, then cos⁡12α=\cos\frac{1}{2}\alpha =

A

−110\frac{- 1}{\sqrt{10}}

B

110\frac{1}{\sqrt{10}}

C

310\frac{3}{\sqrt{10}}

D

−310\frac{- 3}{\sqrt{10}}

Answer

−110\frac{- 1}{\sqrt{10}}

Explanation

Solution

cos⁡(α/2)=−1+cos⁡α2\cos(\alpha/2) = - \sqrt{\frac{1 + \cos\alpha}{2}}

cos⁡α=−1−sin⁡2α\cos\alpha = - \sqrt{1 - \sin^{2}\alpha} [∵α\because\alphalies in IIIrd Quadrant]

=−1−925=−45= - \sqrt{1 - \frac{9}{25}} = - \frac{4}{5}

∴cos⁡(α/2)=−1−452=−110\therefore\cos(\alpha/2) = - \sqrt{\frac{1 - \frac{4}{5}}{2}} = - \frac{1}{\sqrt{10}}.