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If $h(x) = \cos \ x, f(x) = \sin \ x$, and $h = f \circ g$, what is $g$?

$\cos \ x = \sin \ (g(x))$

$g(x) = \sin^{-1} (\cos x)$

How would you proceed from here?

The answer choices are

a) $\frac{\pi}{2} - x$

b) $-x$

c) $\pi - x$

d) $\frac{3\pi}{2} + x$

e) $\frac{3\pi}{2} -x$

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    Can you see that $-sin(x) = -sin(x) * 1 +cos(x) *0$?And how will you continue from here,using basic trigonometric identities? – 2017-02-13
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    Oh shoot. I could have just used the complementary angle relationship between sin and cos angles when cos x = sin g(x). @TakaTiki But thanks anyway for your input. I found it refreshing. – 2017-02-13

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You are lazy !

Compute:

  1. $\sin(\frac{\pi}{2}-x)$

  2. $\sin(-x)$

  3. $\sin(\pi+x)$

  4. $\sin(\frac{3 \pi}{2}+x)$

  5. $\sin(\frac{3 \pi}{2}-x)$.

Then look were the result is $= \cos x$.

For 1., 3., 4. and 5. use the addition theorem.

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    I had already tried your method (and graphing them as well). What I am asking for is a purely algebraic way to solve this so that I can generalize it to other questions. – 2017-02-13