Question

9%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%281%20-%20%20%5Csin%282%20%5Calpha%20%29%20%29" id="TexFormula1" title="( \sin^{2} ( \frac{\pi}{ 4 } - \alpha ) ) = \frac{1}{2} (1 - \sin(2 \alpha ) )" alt="( \sin^{2} ( \frac{\pi}{ 4 } - \alpha ) ) = \frac{1}{2} (1 - \sin(2 \alpha ) )" align="absmiddle" class="latex-formula">
Prove​

Answer 1

Step-by-step explanation:

$\sin^2 (\frac{\pi}{4} - \alpha) = \frac{1}{2}(1 - \sin 2\alpha)$

Use the identity

$\sin^2 \theta = \dfrac{1 - \cos 2\theta}{2}$

on the left side.

$\dfrac{1 - \cos [2(\frac{\pi}{4} - \alpha)]}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\dfrac{1 - \cos (\frac{\pi}{2} - 2\alpha)}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

Now use the identity

$\sin \theta = \cos(\frac{\pi}{2} - \theta)$

on the left side.

$\dfrac{1 - \sin 2\alpha}{2} = \frac{1}{2}(1 - \sin 2\alpha)$

$\frac{1}{2}(1 - \sin 2\alpha) = \frac{1}{2}(1 - \sin 2\alpha)$