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$\begin{array}{l} \cos {75^o}\sin {15^o} = \dfrac{1}{2}\left[ {\sin \left( {{{75}^o} + {{15}^o}} \right) - \sin \left( {{{75}^o} - {{15}^o}} \right)} \right]\\  = \dfrac{1}{2}\left( {\sin {{90}^o} - \sin {{60}^o}} \right) = \dfrac{1}{2}\left( {1 - \dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{{2 - \sqrt 3 }}{4} \end{array}$

Vd6

a)$\begin{array}{l} \sin a\sin \left( {b - c} \right) = \sin a\left( {\sin b\cos c - \sin c\cos b} \right)\\ \sin b\sin \left( {c - a} \right) = \sin b\left( {\sin c\cos a - \sin a\cos c} \right)\\ \sin c\sin \left( {a - b} \right) = \sin c\left( {\sin a\cos b - \sin b\cos a} \right)\\  \Rightarrow A = 0 \end{array}$

b)

$\begin{array}{l} 4\cos x\cos \left( {\dfrac{\pi }{3} - x} \right)\cos \left( {\dfrac{\pi }{3} + x} \right)\\  = 2\cos x.\left[ {\cos \left( {\dfrac{\pi }{3} - x + \dfrac{\pi }{3} + x} \right) + \cos \left( {\dfrac{\pi }{3} + x - \dfrac{\pi }{3} + x} \right)} \right]\\  = 2\cos x\left( {\cos \dfrac{{2\pi }}{3} + \cos 2x} \right)\\  = 2\cos x\cos \dfrac{{2\pi }}{3} + 2\cos 2x\cos x\\  =  - \cos x + 2\cos 2x\cos x\\  =  - \cos x + 2.\dfrac{1}{2}\left( {\cos 3x + \cos x} \right)\\  = \cos 3x \end{array}$