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

1 Answer
Lovecraft
Oct 2, 2015

#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.

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