How do you simplify #sin^2(pi/2-x)-2sin^2x+1#?

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Answer 1 · 2016-11-12T01:53:46

Expand:

#=sin(pi/2 - x) xx sin(pi/2 - x) - 2sin^2x + 1#

Expand #sin(pi/2 - x)# using the formula #sin(A - B) = sinAcosB - cosAsinB#.

#=(sin(pi/2)cosx - cos(pi/2)sinx )(sin(pi/2)cosx - cos(pi/2)sinx)- 2sin^2x + 1#

Evaluate #sin(pi/2)# and #cos(pi/2)#:

#=(1(cosx) - 0(sinx))(1(cosx) - 0(sinx)) - 2sin^2x + 1#

#=(cosx)(cosx) - 2sin^2x + 1#

#=cos^2x - 2(1 - cos^2x) + 1#

#=cos^2x- 2 + 2cos^2x + 1#

#=3cos^2x - 1#

Hopefully this helps!