How do you prove (1+cosx)(1-cosx)=sin^2x?

Mar 14, 2018

${\sin}^{2} x + {\cos}^{2} x = 1$

Explanation:

the identity known is ${\sin}^{2} x + {\cos}^{2} x = 1$.

this can be rearranged to give $1 - {\cos}^{2} x = {\sin}^{2} x$.

using the 'difference of two squares' identity,

where $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$,

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

${1}^{2} = 1$

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

since $1 - {\cos}^{2} x = {\sin}^{2} x$, $\left(1 + \cos x\right) \left(1 - \cos x\right) = {\sin}^{2} x$.

Mar 14, 2018

FOIL the LHS and use the trig identity that ${\sin}^{2} x + {\cos}^{2} x = 1$

Explanation:

First, FOIL the left hand side (LHS):

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

$= 1 - {\cos}^{2} x$

Now, re-write the equation:

$1 - {\cos}^{2} x = {\sin}^{2} x$

There is a trig identity that states:

${\sin}^{2} x + {\cos}^{2} x = 1$

Substitute the 1 in our proof:

${\sin}^{2} x \cancel{+ {\cos}^{2} x - {\cos}^{2} x} = {\sin}^{2} x$

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