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

1 Answer
Oct 13, 2016

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