How do you simplify ${sin}^{2} (\frac{π}{2} - x) - 2 {sin}^{2} x + 1$ $sin^2(pi/2-x)-2sin^2x+1$ ?

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How do you simplify ${sin}^{2} (\frac{π}{2} - x) - 2 {sin}^{2} x + 1$ $sin^2(pi/2-x)-2sin^2x+1$ ?

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Answer 1 · 2016-11-12T01:53:46

Expand:

$= sin (\frac{π}{2} - x) \times sin (\frac{π}{2} - x) - 2 {sin}^{2} x + 1$ $=sin(pi/2 - x) xx sin(pi/2 - x) - 2sin^2x + 1$

Expand $sin (\frac{π}{2} - x)$ $sin(pi/2 - x)$ using the formula $sin (A - B) = sin A cos B - cos A sin B$ $sin(A - B) = sinAcosB - cosAsinB$ .

$= (sin (\frac{π}{2}) cos x - cos (\frac{π}{2}) sin x) (sin (\frac{π}{2}) cos x - cos (\frac{π}{2}) sin x) - 2 {sin}^{2} x + 1$ $=(sin(pi/2)cosx - cos(pi/2)sinx )(sin(pi/2)cosx - cos(pi/2)sinx)- 2sin^2x + 1$

Evaluate $sin (\frac{π}{2})$ $sin(pi/2)$ and $cos (\frac{π}{2})$ $cos(pi/2)$ :

$= (1 (cos x) - 0 (sin x)) (1 (cos x) - 0 (sin x)) - 2 {sin}^{2} x + 1$ $=(1(cosx) - 0(sinx))(1(cosx) - 0(sinx)) - 2sin^2x + 1$

$= (cos x) (cos x) - 2 {sin}^{2} x + 1$ $=(cosx)(cosx) - 2sin^2x + 1$

$= {cos}^{2} x - 2 (1 - {cos}^{2} x) + 1$ $=cos^2x - 2(1 - cos^2x) + 1$

$= {cos}^{2} x - 2 + 2 {cos}^{2} x + 1$ $=cos^2x- 2 + 2cos^2x + 1$

$= 3 {cos}^{2} x - 1$ $=3cos^2x - 1$

Hopefully this helps!

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How do you simplify ${sin}^{2} (\frac{π}{2} - x) - 2 {sin}^{2} x + 1$ $sin^2(pi/2-x)-2sin^2x+1$ ?

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