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`.