How do you simplify #sin(arccos(x))#?

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Answer 1 · 2016-10-21T22:07:41

#sin(arccos(x)) = sqrt(1-x^2)#

From Pythagoras, we have:

#sin^2 theta + cos^2 theta = 1#

If #x in [-1, 1]# and #theta = arccos(x)# then:

#theta in [0, pi]#

#sin(theta) >= 0#

Hence:

#sin(arccos(x)) = sin(theta) = sqrt(1 - cos^2 theta) = sqrt(1-x^2)#

Note we can use the non-negative square root since we have already established that #sin(arccos(x)) >= 0#