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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 20

Answers (1)

Answer:

                \frac{-\pi}{10}

Hint:

You must know the rules of inverse trigonometric function.

Given:

                \sin^{-1}\left ( \cos\frac{33\pi}{5} \right )

Solution:

                \sin^{-1}\left ( \cos\frac{33\pi}{5} \right )

                =\sin^{-1}\left ( \cos\left ( 6\pi+\frac{3\pi}{5} \right ) \right )                                                         \left [ cos(6\pi+\theta) =\cos\theta \right ]

                =\sin^{-1}\left ( \cos\left ( \frac{3\pi}{5} \right ) \right )                                                                       \theta =\sin\left ( \frac{\pi}{2}-\theta \right )

                =\sin^{-1}\left ( \sin\left ( \frac{\pi}{2}-\frac{3\pi}{5} \right ) \right )

                =\sin^{-1}\left ( \sin\left ( \frac{-\pi}{10} \right ) \right )

                = \frac{-\pi}{10}

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