Advertisements
Advertisements
प्रश्न
For the principal value, evaluate of the following:
`sin^-1(-1/2)+2cos^-1(-sqrt3/2)`
Advertisements
उत्तर
`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`
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
