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Proof of product to sum Identities

the product to sum and sum to product identities can be easily derived from the sum and difference identities as follows:

sin(u)sin(v) = ½(cos(u-v) – cos(u+v))

$$ \sin u . \sin v = \dfrac{1}{2}{[\cos(u-v) – \cos(u+v)]} $$

we know that

$$ \cos (u - v) = \cos u \cos v + \sin u \sin v $$ $$ \cos (u + v) = \cos u \cos v - \sin u \sin v $$

by subtracting both side of equation we will get -

$$ \cos (u - v) - \cos (u + v) = \cos u \cos v + \sin u \sin v - ( \cos u \cos v - \sin u \sin v) $$ $$ \cos (u - v) - \cos (u + v) = \cos u \cos v + \sin u \sin v - \cos u \cos v + \sin u \sin v) $$ $$ \cos (u - v) - \cos (u + v) = 2 \sin u \sin v $$

now divide both side by 2

$$ \dfrac{\cos (u - v) - \cos (u + v)}{2} = \dfrac{2 \sin u \sin v}{2} $$ $$ \dfrac{1}{2} [\cos (u - v) - \cos (u + v)] = \sin u \sin v $$

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Proof of sin(u)cos(v) = ½(sin(u+v) + sin(u-v))

$$ \sin u . \cos v = \dfrac{1}{2}{[\sin(u+v) + \sin(u-v)]} $$

we know that

$$ \sin (u + v) = \sin u \cos v + \cos u \sin v $$ $$ \sin (u - v) = \sin u \cos v - \cos u \sin v $$

by adding both side of equation we will get -

$$ \sin (u + v) + \sin (u + v) = \sin u \cos v + \cos u \sin v + \sin u \cos v - \cos u \sin v $$ $$ \sin (u + v)+ \sin (u- v) = 2 \sin u \cos v $$

now divide both side by 2

$$ \dfrac{\sin (u + v) - \sin (u - v)}{2} = \dfrac{2 \sin u \cos v}{2} $$ $$ \dfrac{1}{2} [\sin (u+ v) + \sin (u + v)] = \sin u \cos v $$

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Proof of cos(u)cos(v) = ½(cos(u-v) + cos(u+v))

$$ \cos u . \cos v = \dfrac{1}{2}{[\cos(u-v) + \cos(u+v)]} $$

we know that

$$ \cos (u - v) = \cos u \cos v + \sin u \sin v $$ $$ \cos (u + v) = \cos u \cos v - \sin u \sin v $$

by adding both side of equation we will get -

$$ \cos (u - v) + \cos (u + v) = \cos u \cos v + \sin u \sin v + \cos u \cos v - \sin u \sin v $$ $$ \cos (u - v) + \cos (u + v) = 2 \cos u \cos v $$ now divide both side by 2 $$ \dfrac{\cos (u - v) + \cos (u + v) }{2} = \dfrac{2 \cos u \cos v}{2} $$ $$ \dfrac{1}{2} [\cos (u - v) + \cos (u + v) ] = \cos u \cos v $$

similarly other equation can be proved.