Find the value of \sin ^{-1} [ \sin (-\frac{17 \pi}{8}) ].

CBSE Class 12 Math 2020 Delhi Set 1 Solved Paper - Question 16

Q. Nos. 16 to

\bo {20}

are of very short answer type questions.

Question : 16 of 36

Marks: +1, -0

Find the value of

sin \;^{-1} [ sin \; (-\;{17 π}/{8}) ]

Solution:

sin \;^{-1}( sin \;x)=x ; \;{-π}/{2} ≤ x ≤ \;{π}/{2}

sin \;^{-1} [ sin \; (-\;{17 π}/{8}) ] .

= sin \;^{-1} [- sin \; (\;{17 π}/{8}) ] = sin \;^{-1} [- sin \; (2 π+\;{π}/{8}) ] .

= sin \;^{-1} [- sin \; (\;{17 π}/{8}) ] = sin \;^{-1} [- sin \; (2 π+\;{π}/{8}) ]

= sin \;^{-1} [- sin \; (\;{π}/{8}) ] = sin \;^{-1} [ sin \; (\;{-π}/{8}) ]

=-\;{π}/{8}

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