Trigonometric transformations from sum( or difference) to product and vice-versa

Formulae

sum into product transformations:

2sin A cos B = sin(A+B) + sin(A-B)

2cos A sin B = sin(A+B) - sin(A-B)

2cos A cos B = cos(A-B) + cos(A+B)

2sin A sin B = cos(A-B) - cos(A+B)

product into sum transformations:

sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

sin C - sin D = 2cos( $\frac{C+D}{2}$ ) sin( $\frac{C-D}{2}$ )

cos C + cos D = 2cos( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

cos C - cos D = -2sin( $\frac{C+D}{2}$ ) sin( $\frac{C-D}{2}$ )

Proofs

2sin A cos B = sin(A+B) + sin(A-B)

Proof:

consider Right hand sideo of the statement

sin(A+B) + sin(A-B)

use the coumpund angle formulae i.e. sin(A+B) = sin A cos B + cos A sin B and sin(A-B) = sin A cos B - cos A sin B

sin(A+B) + sin(A-B) = (sin A cos B + cos A sin B) + (sin A cos B - cos A sin B)

sin(A+B) + sin(A-B) =2sin A cos B

hence 2sin A cos B = sin(A+B) + sin(A-B)

sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

Proof:

consider Right hand sideo of the statement

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

this looks liks the previous proof statement i.e 2sin A cos B = sin(A+B) + sin(A-B) where A = $\frac{C+D}{2}$ and B = $\frac{C-D}{2}$

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ ) = sin( $\frac{C+D}{2}$ + $\frac{C-D}{2}$ ) cos ( $\frac{C+D}{2}$ - $\frac{C-D}{2}$ )

2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ ) = sin C + sin D

hence sin C + sin D = 2sin( $\frac{C+D}{2}$ ) cos( $\frac{C-D}{2}$ )

similalry rest of the proofs can be done

2cos A sin B = sin(A+B) - sin(A-B)

Proof:

consider Right hand sideo of the statement

sin(A+B) - sin(A-B)

use the coumpund angle formulae i.e. sin(A+B) = sin A cos B + cos A sin B and sin(A-B) = sin A cos B - cos A sin B

sin(A+B) sin(A-B) = (sin A cos B + cos A sin B) (sin A cos B - cos A sin B)

sin(A+B) + sin(A-B) = 2cos A sin B

hence 2cos A sin B = sin(A+B) - sin(A-B)

2cos A cos B = cos(A-B) + cos(A+B)

proof:

consider Right hand sideo of the statement

cos(A-B) + cos(A+B)

use the coumpund angle formulae i.e. cos(A+B) = cos A cos B - sin A sin B and cos(A-B) = cos A cos B + sin A sin B

cos(A-B) + cos(A+B) = (cos A cos B - sin A sin B) + (cos A cos B - sin A sin B)

cos(A-B) + cos(A+B) = 2cos A cos B

hence 2cos A cos B = cos(A-B) + cos(A+B)

2sin A sin B = cos(A-B) - cos(A+B)

proof:

consider Right hand sideo of the statement

cos(A-B) - cos(A+B)

use the coumpund angle formulae i.e. cos(A+B) = cos A cos B - sin A sin B and cos(A-B) = cos A cos B + sin A sin B

cos(A-B) - cos(A+B) = (cos A cos B - sin A sin B) - (cos A cos B - sin A sin B)

cos(A-B) - cos(A+B) = 2sin A sin B

hence 2sin A sin B = cos(A-B) - cos(A+B)