How do you simplify the expression #cos[arcsin(x)] * sin[arccos(x)]#?

1 Answer
Oct 10, 2016

#cos(arcsin(x))*sin(arccos(x))=1-x^2#

Explanation:

Let's draw a right triangle.

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In this triangle, we've chosen a leg of length #x# and a hypotenuse of length #1#, meaning #sin(a) = cos(b) = x/1 = x#. Because of that, we know that #a = arcsin(x)# and #b = arccos(x)#. Additionally, we can use the Pythagorean theorem to find the length of the remaining leg as #sqrt(1-x^2)#

We can now use this triangle to find the product.

#cos(arcsin(x))*sin(arccos(x)) = cos(a)*sin(b)#

#=sqrt(1-x^2)/1 * sqrt(1-x^2)/1#

#=1-x^2#