How do you find the value of #sin^(-1)(sin (cos^(-1) (sin (pi/12))))#?

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Answer 1 · 2016-07-17T07:36:31

#sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12#.

We will use the following Rules :

#(R1) : sintheta=cos(pi/2-theta)#

#(R2) : cos^-1(costheta)=theta, theta in [o,pi]#

#(R3) : sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]#

Let us note that, by

#(R1), sin(pi/12)=cos(pi/2-pi/12)=cos(5pi/12)#

So, #cos^-1(sin(pi/12))=cos^-1(cos(5pi/12))#, where,

#5pi/12 in [0,pi]#, so, using #(R2)#, we get,

#cos^-1(cos(5pi/12))=5pi/12#

Again, as #5pi/12 in [-pi/2,pi/2]#, by #(R3)#, we have,

#sin^-1(sin(5pi/12))=5pi/12#

Finally, #sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12#.

Hope, this will be of Help! Enjoy Maths.!

Answer 2 · 2016-07-17T07:45:15

#(5pi)/12=75^o#

Use #sin a = cos (pi/2-a) and,#

if # y=f(x) , x = f^(-1)y, f f^(-1)y=y and f^(-1)yf(x)=x#.

The given expression is

#sin^(-1)sin cos^(-1)sin(pi/12)#

#=sin^(-1)sin (cos^(-1)cos(pi/2-pi/12))#

#=sin^(-1)sin((5pi)/12)#

#=(5pi)/12#