Evaluate each of the following:

For finding the solution we first of need to find the principal value of


Sin–1


Let,


Sin–1 =y


sin y =


sin


The range of principal value of sin–1 is and sin


Therefore, the principal value of Sin–1 is


The above equation changes to cot–1(2cos)


Now we need to find the value of 2cos


cos


2cos = 1 x


2cos = 1


Now the equation simplification to cot–1(1)


Let cot–1(1) = y


cot y = 1


= cot = 1


The range of principal value of cot–1is (0, π)


and cot = 1


The principal value of cot–1(2cos(Sin–1)) is


3