Trigonometry/More About Addition Formulas

Related Formulae

$\sin (a-b)=\sin (a)\cos (b)-\cos (a)\sin (b)$

$\sin (2a)=2\sin (a)\cos (a)$

$\sin \left(\frac{a}{2}\right)=\pm \sqrt{\frac{1-\cos (a)}{2}}$

Tangent Formulae

$\tan (a+b)=\frac{\tan (a)+\tan (b)}{1-\tan (a)\tan (b)}$

$\tan (a-b)=\frac{\tan (a)-\tan (b)}{1+\tan (a)\tan (b)}$

$\tan (2a)=\frac{2\tan (a)}{1-{\tan }^2(a)}=\frac{2\cot (a)}{{\cot }^2(a)-1}=\frac{2}{\cot (a)-\tan (a)}$

$\tan \left(\frac{a}{2}\right)=\pm \sqrt{\frac{1-\cos (a)}{1+\cos (a)}}=\frac{\sin (a)}{1+\cos (a)}=\frac{1-\cos (a)}{\sin (a)}=\frac{-1\pm \sqrt{1+{\tan }^2(a)}}{\tan (a)}$

In the last row of expressions, if $0^{\circ }\le a\le 90^{\circ }$ 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 $\sin (a)=\cos (90^{\circ }-a)$ . Use the formula for $\cos (a-b)$ and cofunctions we can write

$\sin (a+b)$	$=\cos (90-(a+b))$
	$=\cos ((90^{\circ }-a)-b)$
	$=\cos (90^{\circ }-a)\cos (b)+\sin (90^{\circ }-a)\sin (b)$
	$=\sin (a)\cos (b)+\cos (a)\sin (b)$

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

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 Formulae

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

$\cos (2a)={\cos }^2(a)-{\sin }^2(a)=2{\cos }^2(a)-1=1-2{\sin }^2(a)$

$\cos \left(\frac{a}{2}\right)=\pm \sqrt{\frac{1+\cos (a)}{2}}$

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

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