Prove that $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$
$\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)$
$=\frac{1}{2}\left[2 \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)\right]+\frac{1}{2}\left[-2 \sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)\right]$
$=\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}+\cos \left\{\left(\frac{\pi}{4}-x\right)-\left(\frac{\pi}{4}-y\right)\right\}\right]$
$+\frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}-\cos \left\{\frac{\pi}{4}-x\right\}-\left(\frac{\pi}{4}-y\right)\right]$
$\left[ \begin{gathered}
\because 2\cos A\cos B = \cos (A + B) + \cos (A - B) \hfill \\
- 2\sin A\sin B = \cos (A + B) - \cos (A - B) \hfill \\
\end{gathered} \right]$
$=2 \times \frac{1}{2}\left[\cos \left\{\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right\}\right]$
$=\cos \left[\frac{\pi}{4}-(x+y)\right]$
$=\sin (x+y)$
$= R . H.S$
Find the radian measures corresponding to the following degree measures:
$-47^{\circ} 30^{\prime}$
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$\sin 75^{\circ}$
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Find the degree measures corresponding to the following radian measures (Use $\pi=\frac{22}{7}$ ).
$-4$
If $a\cos \theta + b\sin \theta = m$ and $a\sin \theta - b\cos \theta = n,$ then ${a^2} + {b^2} = $