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Question

Mathematics Question on Inverse Trigonometric Functions

sin⁡(12cos⁡−145)\sin\left(\frac{1}{2} \cos^{-1} \frac{4}{5}\right) is equal to:

A

−110- \frac{1}{\sqrt{10}}

B

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

C

−110 - \frac{1}{10}

D

110\frac{1}{10}

Answer

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

Explanation

Solution

Given: sin⁡(12cos⁡−145)\sin \left( \frac{1}{2} \cos^{-1} \frac{4}{5}\right) Now, Let 45=cos⁡2θ \frac{4}{5} = \cos 2 \theta ∴sin⁡(12cos⁡−145)=sin⁡(12cos⁡−1cos⁡2θ)\therefore \sin \left( \frac{1}{2} \cos^{-1} \frac{4}{5} \right) = \sin\left(\frac{1}{2} \cos^{-1} \cos 2 \theta \right) =sin⁡(12×2θ)=sin⁡θ=1−cos⁡2θ2= \sin\left( \frac{1}{2} \times2 \theta\right) = \sin\theta = \sqrt{\frac{1- \cos2\theta}{2}} =1−452=110= \sqrt{\frac{1- \frac{4}{5}}{2}} = \frac{1}{\sqrt{10}}