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

2 Answers
May 4, 2018

#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

May 4, 2018

#"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"#