# Factorization Formulas

This is just an extension to previous 2 sections, viz. trigonometric functions and compound angles. We will make use of the formulas learned earlier to get to some new formulas, which are equally useful.

Specifically, the addition and subtraction formulas will now be expressed as product of 2 trigonometric functions. The terms in the product are termed as factors, hence the name factorization formulas.

Let $C$ and $D$ be any 2 angles.

• ### Sum and Difference into Products

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

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

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

${cos(C)- cos(D)= -2 sin ( \frac {C+D}{2} ) sin ( \frac {C-D}{2} )}$. *** Note the minus sign.

Let $A$ and $B$ be any 2 angles.

• ### Product into Sum and Difference

${sin (A+B) + sin (A-B) = 2sin(A)cos(B)}$

${sin (A+B) - sin (A-B) = 2cos(A)sin(B)}$

${cos(A+B)+ cos(A-B) = 2cos(A)cos(B)}$

${cos(A+B)- cos(A-B) = -2sin(A)sin(B)}$,  *** Note the minus sign.

Following are the formulas with $2$ taken to the RHS (in the denominator)

${sin(A)cos(B) = \frac 1 2 [sin(A+B)+ sin(A-B)]}$

${cos(A)sin(B) = \frac 1 2 [sin(A+B)- sin(A-B)]}$

${cos(A)cos(B) = \frac 1 2 [cos(A+B)+cos(A-B)]}$

${sin(A)sin(B)= \frac 1 2 [cos(A-B)-cos(A+B)]}$

• ### Application to Angles of a Triangle

Consider $\Delta ABC$. We know that $\angle A + \angle B + \angle C = 180^o = \pi^c$.

${sin (A+B) = sin (\pi -C) = sin (C)}$

${sin (B+C) = sin (\pi -A) = sin (A)}$

${sin (A+C) = sin (\pi -B) = sin (B)}$

${cos (A+B) = cos (\pi - C) = -cos (C)}$

${cos (B+C) = cos (\pi - A) = -cos (A)}$

${cos (A+C) = cos (\pi - B) = -cos (B)}$

${sin \{ \frac {A+B}{2} \} = sin \{ \frac {\pi}{2} - \frac {C}{2} \} = cos \{ \frac C 2 \}}$

${sin \{ \frac {B+C}{2} \} = sin \{ \frac {\pi}{2} - \frac {A}{2} \} = cos \{ \frac A 2 \}}$

${sin \{ \frac {A+C}{2} \} = sin \{ \frac {\pi}{2} - \frac {B}{2} \} = cos \{ \frac B 2 \}}$

${cos \{ \frac {A+B}{2} \} = cos \{ \frac {\pi}{2} - \frac {C}{2} \} = sin \{ \frac C 2 \}}$

${cos \{ \frac {B+C}{2} \} = cos \{ \frac {\pi}{2} - \frac {A}{2} \} = sin \{ \frac A 2 \}}$

${cos \{ \frac {A+C}{2} \} = cos \{ \frac {\pi}{2} - \frac {B}{2} \} = sin \{ \frac B 2 \}}$