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#