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  • State True or False for the statement The principal value of sin inverse into bracket cos into bracket sin inverse 1 divided 2 is pi divided 3

State True or False for the statement

The principal value of \sin^{-1}\left [ \cos\left ( \sin^{-1}\frac{1}{2} \right ) \right ] is \frac{\pi}{3}

Answers (1)

True

Principal value of sin-1 x is \left [ -\frac{\pi}{2},\frac{\pi}{2} \right ]

Principal value of cos-1 x is [0, π]

We have, \sin^{-1}\left [ \cos \left [ \sin^{-1}\left (\frac{1}{2} \right ) \right ] \right ]

As, \sin\frac{\pi}{6}=\frac{1}{2}  so

\sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \cos\left [ \sin^{-1}\left (\sin \frac{\pi}{6} \right ) \right ] \right ]

\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \cos \left [ \frac{\pi}{6} \right ]\right ]

As, \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} so,

\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\sin^{-1}\left [ \sin \left [ \frac{\pi}{3} \right ]\right ]

\Rightarrow \sin^{-1}\left [ \cos\left [ \sin^{-1}\left ( \frac{1}{2} \right ) \right ] \right ]=\frac{\pi}{3}

 

 

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