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,

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

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