How do you prove $sin x = 1 - 2 {cos}^{2} x$ $sin x = 1 - 2 cos^2x$ ?

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How do you prove $sin x = 1 - 2 {cos}^{2} x$ $sin x = 1 - 2 cos^2x$ ?

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Answer 1 · 2015-05-16T23:44:49

For $x = \frac{π}{6}$ $x=pi/6$ :
$sin (x) = \frac{1}{2}, cos (x) = \frac{\sqrt{3}}{2}$ $sin(x) = 1/2, cos(x) = sqrt(3)/2$
$1 - 2 \cdot {(\frac{\sqrt{3}}{2})}^{2} = 1 - 2 \cdot \frac{3}{4} = 1 - \frac{3}{2} = - \frac{1}{2} \neq \frac{1}{2}$ $1 - 2 * (sqrt(3)/2)^2 = 1-2*3/4 = 1-3/2 = -1/2 != 1/2$ //

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How do you prove $sin x = 1 - 2 {cos}^{2} x$ $sin x = 1 - 2 cos^2x$ ?

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