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Question

Mathematics Question on Properties of Inverse Trigonometric Functions

The value of \cos \left\\{ \sin^{-1} \frac{3}{5} \right\\} is :

A

35\frac{3}{5}

B

45\frac{4}{5}

C

−46- \frac{4}{6}

D

−53- \frac{5}{3}

Answer

45\frac{4}{5}

Explanation

Solution

Let sin⁡−1(35)=θ⇒35=sin⁡θ\sin^{-1} \left(\frac{3}{5}\right) = \theta \Rightarrow \frac{3}{5} = \sin\theta We know cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1- \sin^{2} \theta} =1−925=1625=45= \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} ∴cos⁡[sin⁡−1(35)]=cos⁡θ=45\therefore \cos\left[\sin^{-1} \left(\frac{3}{5}\right) \right] = \cos\theta = \frac{4}{5}