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

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Answer 1 · 2018-03-29T02:43:05

#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#

#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#