The value of $sin[\frac{π}{2}-sin^{-1

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

The value of $sin[\frac{π}{2}-sin^{-1}(-\frac{\sqrt{3}}{2})]$ is:

Options:

$-\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\frac{\sqrt{3}}{2}$

$-\frac{\sqrt{3}}{2}$

Correct Answer:

$-\frac{1}{2}$

Explanation:

$sin[\frac{π}{2}-sin^{-1}(-\frac{\sqrt{3}}{2})]$

$=sin[\frac{π}{2}-sin^{-1}(sin(\frac{4π}{3}))]$

$=sin[\frac{π}{2}-\frac{4π}{3}]$

$=sin[\frac{3π-8π}{6}]$

$=sin[\frac{-5π}{6}]$

$=-sin[\frac{5π}{6}]$

$=-\frac{1}{2}$