proof of double angle identity

Sine:

$\\displaystyle\\sin(2a)$	$\\displaystyle=$	$\\displaystyle\\sin(a+a)$
	$\\displaystyle=$	$\\displaystyle\\sin(a)\\cos(a)+\\cos(a)\\sin(a)$
	$\\displaystyle=$	$\\displaystyle 2\\sin(a)\\cos(a).$

Cosine:

$\\displaystyle\\cos(2a)$	$\\displaystyle=$	$\\displaystyle\\cos(a+a)$
	$\\displaystyle=$	$\\displaystyle\\cos(a)\\cos(a)+\\sin(a)\\sin(a)$
	$\\displaystyle=$	$\\displaystyle\\cos^{2}(a)-\\sin^{2}(a).$

By using the identity

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

we can change the expression above into the alternate forms

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

Tangent:

$\\displaystyle\\tan(2a)$	$\\displaystyle=$	$\\displaystyle\\tan(a+a)$
	$\\displaystyle=$	$\\displaystyle\\frac{\\tan(a)+\\tan(a)}{1-\\tan(a)\\tan(a)}$
	$\\displaystyle=$	$\\displaystyle\\frac{2\\tan(a)}{1-\\tan^{2}(a)}.$