# How do you verify (1 + cos theta)(1 - cos theta) = sin^2 theta?

##### 2 Answers
May 25, 2015

Use Pythagoras and the basic definitions of sine and cos
sin x = $\frac{o}{h}$
cos x = $\frac{a}{h}$
where o =the side opposite the angle x, a = the side adjacent to the angle x and h equals the hypotenuse of the right-angled triangle.

Pythagora states ${h}^{2} = {a}^{2} + {o}^{2}$
therefore $1 = {a}^{2} / {h}^{2} + {o}^{2} / {h}^{2}$
so 1 = ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)$
${\sin}^{2} \left(x\right) = 1 - {\cos}^{2} \left(x\right)$
tthen ${\sin}^{2} \left(x\right) = \left(1 - \cos \left(x\right)\right) \left(1 + \cos \left(x\right)\right)$

May 26, 2015

$\left(1 + \cos \theta\right) \left(1 - \cos \theta\right) = 1 - {\cos}^{2} \theta$

$= \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) - {\cos}^{2} \theta$

$= {\sin}^{2} \theta + \left({\cos}^{2} \theta - {\cos}^{2} \theta\right)$

$= {\sin}^{2} \theta$