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

Question

Josiah Anderson · Accepted Answer

#(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#

Charles Autterson · Answer

#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.!

Dylan Arnold · Answer

undefined To find the value of $ \sin^{-1}(\sin(\cos^{-1}(\sin(\frac{\pi}{12})))) $, first evaluate the innermost function:

1. $ \cos^{-1}(\sin(\frac{\pi}{12})) $: Use the fact that $ \sin(\frac{\pi}{12}) = \sin(15^\circ) $. Since $ \cos(x) = \sin(90^\circ - x) $, $ \cos^{-1}(\sin(\frac{\pi}{12})) = \cos^{-1}(\sin(15^\circ)) = 75^\circ $.

2. $ \sin^{-1}(\cos^{-1}(\sin(\frac{\pi}{12}))) $: Now, $ \sin^{-1}(\cos^{-1}(0.258819)) $, and $ \sin^{-1}(\cos^{-1}(0.258819)) = 15^\circ $.

So, $ \sin^{-1}(\sin(\cos^{-1}(\sin(\frac{\pi}{12})))) = 15^\circ $.