# Transforming Expressions that contain Inverse trigonometric Functions

To transform expressions that contain inverse trigonometric function, we use definitions of these functions and trigonometric formulas.

Example 1. Simplify sin(arccos(x)), where -1<=x<=1.

Let y=arccos(x). Then, according to definition of inverse cosine, cos(y)=x and 0<=y<=pi.

We need to find sin(y).

It is known that sin^2(y)=1-cos^2(y). Therefore, sin^2(y)=1-x^2. This means that either sin(y)=sqrt(1-x^2) or sin(y)=-sqrt(1-x^2). But 0<=y<=pi and on segment [0,pi] sine takes only non-negative values. Therefore, sin(y)=sqrt(1-x^2).

So, sin(arccos(x))=sqrt(1-x^2).

Example 2. Calculate tan(1/2arccos(-3/5)).

Let alpha=arccos(-3/5). Then, according to definition of inverse cosine, cos(alpha)=-3/5 and pi/2<alpha<pi (this is subinterval of interval [0,pi] where cosine is negative).

We need to find tan(alpha/2).

We have that cos^2(alpha/2)=(1+cos(alpha))/2=(1+(-3/5))/2=1/5.

Next, since 1+tan^2(alpha/2)=1/(cos^2(alpha/2)) then tan^2(alpha/2)=1/(cos^2(alpha/2))-1=1/(1/5)-1=5-1=4.

This means that either tan(alpha/2)=sqrt(4)=2 or tan(alpha/2)=-sqrt(4)=-2.

But we have that pi/2<alpha/pi, i.e. pi/4<alpha/2<pi/2. On interval (pi/4,pi/2) tangent is positive, i.e. tan(alpha/2)>0. Therefore, tan(alpha/2)=2, i.e. tan(1/2arccos(-3/5))=2.