Question

For the principal value, evaluate of the following:

`sin^-1(-1/2)+2cos^-1(-sqrt3/2)`

Answer 1

`sin^-1(-1/2)+2cos^-1(-sqrt3/2)`

`=sin^-1{sin(-pi/6)}+2cos^-1(cos  (5pi)/6)`  `[because "Range of sine is"[-pi/2,pi/2];-pi/6in[-pi/2,pi/2] "and range of cosine is"  [0,pi]  ;  (5pi)/6 in[0,pi]]` 

`=-pi/6+2((5pi)/6)`

`=-pi/6+(5pi)/3`

`=(9pi)/6`

`=(3pi)/2`

`therefore sin^-1(-1/2)+2cos^-1(-sqrt3/2)=(3pi)/2`