Question

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

Answer 1

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

Explanation:

Let's draw a right triangle.

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`