Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
Question.

Prove that: $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$


solution:

$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$

$=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]$

$=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$

$+\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$

$\left[\begin{array}{l}\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B) \\ -2 \sin A \sin B=\cos (A+B)-\cos (A-B)\end{array}\right]$

$=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]$

$=\cos \left[\frac{\pi}{2}-(x+y)\right]$

$=\sin (x+y)$

$=\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S}$
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