How do you evaluate #sin^-1(sin((11pi)/10))#?

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Answer 1 · 2018-03-29T20:08:58

Evalute the inner bracket first. See below.

#sin(11*pi/10) = sin((10+1)pi/10 =sin(pi + pi/10)#

Now use the identity:

#sin(A+B) = sinAcosB +cosAsinB#

I leave the nitty-gritty substitution for you to solve.

Answer 2 · 2018-03-29T20:38:46

#sin^-1(sin((11pi)/10))=-pi/10#

Note:

#color(red)((1)sin(pi+theta)=-sintheta#

#color(red)((2)sin^-1(-x)=-sin^-1x#

#color(red)((3)sin^-1(sintheta)=theta,where,theta in[-pi/2,pi/2]#

WE have,

#sin^-1(sin((11pi)/10))=sin^-1(sin((10pi+pi)/10))#

#=sin^-1(sin(pi+pi/10)).........toApply (1)#

#=sin^-1(-sin(pi/10))...........toApply(2)#

#=-sin^-1(sin(pi/10))..........toApply(3)#

#=-pi/10 in [-pi/2,pi/2]#

Hence,

#sin^-1(sin((11pi)/10))=-pi/10#