Question

How do you evaluate `sin(2arccos(3/5))`?

Answer 1

It's `24/25`

Explanation:

We need to make use of the property that `cos(arccos(x))=x`. At the moment we have a sine instead of a cosine. We also know the formula: `cos^2x+sin^2x=1`.
To make use of both of these, we need to square the expression and immediately take the square root (so we don't do anything illegal):

`sqrt(sin^2(2arccos(3/5)))=sqrt(1-cos^2(2arccos(3/5))`

The only problem that we have now is the `2` in front of the `arccos`. We can solve this by ussing the double angle formula:

`cos(2a)=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a`

We need it in therms of the cosine, so let's take the second one:

`sqrt(1-(2cos^2(arccos(3/5))-1)^2`
Simplifying this:
`sqrt(1-4*cos^4(arccos(3/5))+4cos^2(arccos(3/5))-1)`
Now we can replace `cos(arccos(x))` by `x`:
`sqrt(4*(3/5)^2-4*(3/5)^4)=sqrt(4*(3/5)^2*(1-(3/5)^2)`
`=2*3/5*sqrt(1-9/25)=6/5*sqrt(16/25)=6/5*4/5=24/25`