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!

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