Cos(2sin^-1 x -pi/2) ?

1 Answer
Ratnaker Mehta
Jan 30, 2018

# 2xsqrt(1-x^2), |x| le 1#.

Explanation:

Recall the definition of #sin^-1# function :

#sin^-1x=theta, |x|le1 iff sintheta=x, theta in [-pi/2,pi/2]#.

Then, #cos(2sin^_1x-pi/2)#,

#=cos(2theta-pi/2)=cos(pi/2-2theta)#,

#=sin2theta#,

#=2sinthetacostheta#,

#=2sinthetasqrt(1-sin^2theta)#.

# rArr cos(2sin^-1x-pi/2)=2xsqrt(1-x^2), |x| le 1#.

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