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

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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)

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Marcel Gleason
15.5k 3 10 26
answered 6 months ago