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

May 4, 2018

$2 = 2$

#### Explanation:

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

$\textcolor{red}{{\sin}^{2} x} - 2 \sin x \cos x + \textcolor{red}{{\cos}^{2} x} + \textcolor{b l u e}{{\sin}^{2} x} + 2 \sin x \cos x + \textcolor{b l u e}{{\cos}^{2} x} = 2$

red terms equal 1
from the Pythagorean theorem
also, blue terms equal 1

So

$1 \textcolor{g r e e n}{- 2 \sin x \cos x} + 1 \textcolor{g r e e n}{+ 2 \sin x \cos x} = 2$

green terms together equal 0

So now you have

$1 + 1 = 2$

$2 = 2$

True

May 4, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

â€¢color(white)(x)sin^2x+cos^2x=1

$\text{consider left side}$

$\text{expand each factor using FOIL}$

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

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

$\text{adding the right sides gives}$

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

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

$= 2 \times 1 = 2 = \text{ right side "rArr"proven}$