# Formula and Proof of product to sum Identities

### Formulas:

sin(u)sin(v) = ½(cos(u-v) – cos(u+v))

sin(u)cos(v) = ½(sin(u+v) + sin(u-v))

cos(u)sin(v) = ½(sin(u+v) – sin(u-v))

cos(u)cos(v) = ½(cos(u+v) + cos(u-v))

### 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

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

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

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

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.