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Convert the following product into the sum or difference of sine and cosine:
\(\sin \frac{5 \pi}{12} \cos \frac{\pi}{12}\)
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Convert the following product into the sum or difference of sine and cosine:
\(\sin \frac{5 \pi}{12} \cos \frac{\pi}{12}\)

Answer

\(\sin \frac{5 \pi}{12} \cos \frac{\pi}{12}=\frac{1}{2}\left(2 \sin \frac{5 \pi}{12} \cos \frac{\pi}{12}\right)\)
\(=\frac{1}{2}\left\{\sin \left(\frac{5 \pi}{12}+\frac{\pi}{12}\right)+\sin \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)\right\}\) \(2 \sin x \cos y = \sin (x+y) + \sin (x-y)\)\(=\frac{1}{2}\left(\sin \frac{\pi}{2}+\sin \frac{\pi}{3}\right)\)
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Prove that: \(4\cos 12^{\circ }\cos 48^{\circ }\cos 72^{\circ }=\cos 36^{\circ}\)
Prove that: \(\sin A\sin (\frac {\pi }{3}-A)\sin (\frac {\pi }{3}+A)=\frac {1}{4}\sin 3A\)
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