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Question: \(\cos^{2}\left( \frac{\pi}{4} - \beta \right) - \sin^{2}\left( \alpha - \frac{\pi}{4} \right) =\)...

cos⁡2(π4−β)−sin⁡2(α−π4)=\cos^{2}\left( \frac{\pi}{4} - \beta \right) - \sin^{2}\left( \alpha - \frac{\pi}{4} \right) =

A

sin⁡(α+β)sin⁡(α−β)\sin(\alpha + \beta)\sin(\alpha - \beta)

B

cos⁡(α+β)cos⁡(α−β)\cos(\alpha + \beta)\cos(\alpha - \beta)

C

sin⁡(α−β)cos⁡(α+β)\sin(\alpha - \beta)\cos(\alpha + \beta)

D

sin⁡(α+β)cos⁡(α−β)\sin(\alpha + \beta)\cos(\alpha - \beta)

Answer

sin⁡(α+β)cos⁡(α−β)\sin(\alpha + \beta)\cos(\alpha - \beta)

Explanation

Solution

cos⁡2(π4−β)−sin⁡2(α−π4)\cos^{2}\left( \frac{\pi}{4} - \beta \right) - \sin^{2}\left( \alpha - \frac{\pi}{4} \right)

=cos⁡(π4−β+α−π4)cos⁡(π4−β−α+π4)= \cos\left( \frac{\pi}{4} - \beta + \alpha - \frac{\pi}{4} \right)\cos\left( \frac{\pi}{4} - \beta - \alpha + \frac{\pi}{4} \right)

=cos⁡(α−β)cos⁡(π2−α+β‾)=cos⁡(α−β)sin⁡(α+β)= \cos(\alpha - \beta)\cos\left( \frac{\pi}{2} - \overline{\alpha + \beta} \right) = \cos(\alpha - \beta)\sin(\alpha + \beta).