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Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 1 Subquestion (ii) Maths Textbook Solution.

Answers (1)

Answer:
\frac{-\pi }{6}
Hint:
\cos \left ( 90^{\circ}+x \right )= -\sin x
Given:
Find\: principal\: value\: o\! f \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )
Solution:
Let, \sin^{-1}\left ( \cos \frac{2\pi }{3} \right )= y
\Rightarrow \sin y= \left ( \cos \frac{2\pi }{3} \right )
\Rightarrow \sin y=\cos \frac{2\pi }{3}= \cos \left ( \frac{\pi }{2}+\frac{\pi }{6} \right )= -\sin\left ( \frac{\pi }{6} \right )
W\! e\, know\, principal\, value\, o\! f \sin^{-1} is\, \left [ \frac{-\pi }{2},\frac{\pi }{2} \right ] and -\sin \sin \left ( \frac{\pi }{6} \right )= \cos \cos \left ( \frac{2\pi }{3} \right )
Therefore, principal\, value\, o\! f \, \sin^{-1}\left [ \cos \left ( \frac{2\pi }{3} \right ) \right ] is\, \frac{-\pi }{6}.

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