Question

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

Answer 1

`2=2`

Explanation:

`(sinx-cosx)^2+(sinx+cosx)^2 = 2`

`color(red)(sin^2x) - 2 sinx cosx +color(red)(cos^2x) + color(blue)(sin^2x) + 2 sinx cosx +color(blue)(cos^2x) = 2`

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

So

`1 color(green)(- 2 sinx cosx) + 1 color(green)(+2 sinx cosx) = 2`

green terms together equal 0

So now you have

`1 + 1 = 2`

`2 = 2`

True

Answer 2

`"see explanation"`

Explanation:

`"using the "color(blue)"trigonometric identity"`

`•color(white)(x)sin^2x+cos^2x=1`

`"consider left side"`

`"expand each factor using FOIL"`

`(sinx-cosx)^2=sin^2xcancel(-2cosxsinx)+cos^2x`

`(sinx+cosx)^2=sin^2xcancel(+2cosxsinx)+cos^2x`

`"adding the right sides gives"`

`2sin^2x+2cos^2x`

`=2(sin^2x+cos^2x)`

`=2xx1=2=" right side "rArr"proven"`