Question

@TuringTest

Answer 1

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Answer 2

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

Answer 3

\[sin^{-1}\left[cos\left(-\frac{\pi}{10}\right)\right]\]

Answer 4

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

Answer 5

\[
\sin(\sin^{-1}(x)) = x
\]\[
\cos(\sin^{-1}(x)) = \sqrt{\cos^2(\sin^{-1}(x))} = \sqrt{1-\sin^2(\sin^{-1}(x))} = \sqrt{1-x^2}
\]

Answer 6

\[cos^{-1}\left(sin\frac{7\pi}{6}\right)\]

Answer 7

By symmetry: \[
\sin(\cos^{-1}(x)) = \sqrt{1-x^2}
\]

Answer 8

\[cos^{-1}-\frac 12 \]

Answer 9

\[cos^{-1}\left(-\frac 12\right)\]

Answer 10

If we restrict the domain of \(y=cosx\) to the interval \((0,\pi)\) to the interval \([0,\pi]\), the restricted function
\[y=cosx \;\;\;o\le x\le \pi\]
is one to one and hence will have an inverse function.

Answer 11

\[sec\left(tan^{-1}\frac 12\right)\]