Question

How do you calculate `cos(tan^-1(4/3)-sin^-1(12/13))`?

Answer 1

`cos(tan^-1(4/3)-sin^-1(12/13))=63/65`

Explanation:

`cos(tan^-1(4/3)-sin^-1(12/13))`
The restriction for the range of `tan^-1x` is `(-pi/2,pi/2)` and the restriction for `sin^-1x` is`[-pi/2,pi/2]`.

We need to draw two triangles. They will both be in quadrant I since both the arguments are positive. Let's call the first one triangle A and the second one triangle B.

For triangle A the opposite is 4 and adjacent is 3 so the hypotenuse is 5. For triangle B the opposite is 12 and the hypotenuse is 13 so the adjacent is 5. Now

Use the formula `cos(A-B)=cos Acos B+sin A sin B` to evaluate. Note that `A=tan^-1(4/3)` and `B=sin^-1(12/13)`. Therefore,

`cos(tan^-1(4/3)-sin^-1(12/13))`

`=cos(tan^-1(4/3))cos(sin^-1(12/13))+sin (tan^-1(4/3)) sin (sin^-1(12/13))`

`=cos Acos B+sin A sin B`-> Use triangles A and B to find the ratios

`=3/5 * 5/13+ 4/5 * 12/13`

`=3/13+48/65=63/65`