How do you find the exact value for sin^-1 (cos (2pi/3))?

1 Answer
Mar 29, 2018

sin^-1(cos(2pi/3))=7pi/6, 11pi/6
Among which the first positive solution happens to be

sin^-1(cos(2pi/3))=7pi/6

Explanation:

sin^-1(cos(2pi/3))=?

2pi/3=pi-pi/3

cos(2pi/3)=cos(pi-pi/3)

cos(A-B)=cosAcosB+sinAsinB

cos(pi-pi/3)=cospicos(pi/3)+sinpisin(pi/3)

cospi=-1

cos(pi/3)=1/2

sinpi=0

sin(pi/3)=sqrt3/2

cos(pi-pi/3)=-1xx1/2+0xxsqrt3/2

cos(2pi/3)=-1/2

sin^-1(cos(2pi/3))=sin^-1(-1/2)

Let

u=sin^-1(-1/2)

sinu=-1/2

sin(pi/6)=1/2
sine ratio is negative in 3rd and 4th quadrants

For angles u<2pi,

sin(pi+pi/6)=-sin(pi/6)->

u=pi+pi/6=7pi/6

sin(2pi-pi/6)=-sin(pi/6)->

u=2pi-pi/6=11pi/6

Hence,
Exact value of

sin^-1(cos(2pi/3))=7pi/6, 11pi/6