Question

What is the value of `sin -1 (cos x)`?

Answer 1

`arcsin(cos(x)) = x+pi/2`

Explanation:

Assuming you mistyped and meant `sin^(-1)(cos(x))` or simply `arcsin(cos(x))`, we can easily solve this by putting it on terms of the sine function.

We know that the cosine function, is nothing more than the sine `pi/2` radians out of phase, as proved below:

`cos(theta-pi/2) = cos(theta)cos(-pi/2) - sin(theta)sin(-pi/2)`
`cos(theta-pi/2) = cos(theta)*0 -(-sin(theta)sin(pi/2))`
`cos(theta-pi/2) = sin(theta)*1 = sin(theta)`

So we can say that the sine function, 90 degrees ahead, is the cosine function.

`arcsin(cos(x)) = arcsin(sin(x+pi/2))`

Using the property of inverse functions that `f^(-1)(f(x)) = x`, we have

`arcsin(cos(x)) = x+pi/2`

If you must use degrees, just convert those `pi/2` radians to `90º` degrees.