## Related Formulas

$\displaystyle \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

$\sin 2a = 2 \sin a \cos a\,$

$\sin \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{2}} \,$

## Tangent Formulas

$\tan(a+b)= \frac{\tan(a) + \tan(b)}{1-\tan(a)\tan(b)}$
$\tan(a-b)= \frac{\tan(a) - \tan(b)}{1+\tan(a)\tan(b)}$
$\tan 2A = {2 \tan A \over 1 - \tan^2 A} = {2 \cot A \over \cot^2 A - 1} = {2 \over \cot A - \tan A} \,$
$\tan \frac{A}{2} = \pm \sqrt{\frac{1-\cos A}{1+\cos A}} = \frac {\sin A}{1+\cos A} = \frac {1-\cos A}{\sin A} = \frac {-1 \pm \sqrt {1+\tan^2 A}}{\tan A}$

In the last row of expressions, if A is between 0º and 90º then the trigonometric functions are all positive so the positive sign is needed before the square root.

## Derivations

• sin(a + b) = sin a cos b + cos a sin b

Using cofunctions we know that sin a = cos (90 - a). Use the formula for cos(a - b) and cofunctions we can write

         sin(a + b) = cos(90 - (a + b))
= cos((90 - a) - b)
= cos(90 -a)cos b + sin(90 - a)sin b
= sin a cos b + cos a sin b

• sin(a - b) = sin a cos b - cos a sin b

Having derived sin(a + b) we replace b with "-b" and use the fact that cosine is even and sine is odd.

      sin(a + (-b)) = sin a cos (-b) + cos a sin (-b)
= sin a cos b + cos a (-sin b)
= sin a cos b - cos a sin b


## Related Formulas

$\cos (A - B) = \cos A \cos B + \sin A \sin B \,$
$\cos 2A = \cos^2 A - \sin^2 A = 2 \cos^2 A -1 = 1-2 \sin^2 A \,$
$\cos \frac{A}{2} = \pm \sqrt{\frac{1+\cos A}{2}} \,$

## Derivations

• cos(a + b) = cos a cos b - sin a sin b
• cos(a - b) = cos a cos b + sin a sin b

Using cos(a + b) and the fact that cosine is even and sine is odd, we have

            cos(a + (-b)) = cos a cos (-b) - sin a sin (-b)
= cos a cos b - sin a (-sin b)
= cos a cos b + sin a sin b