How do you use the double angle or half angle formulas to simplify #(sinx-cosx)^2#?

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Answer 1 · 2015-10-26T01:32:30

#1-sin(2x)#

Expand.

#[1]" "(sinx-cosx)^2#

#[2]" "=sin^2x-2sinxcosx+cos^2x#

Pythagorean Identity: #sin^2alpha+cos^2alpha=1#

#[3]" "=(sin^2x+cos^2x)-2sinxcosx#

#[4]" "=1-2sinxcosx#

Double Angle Identity: #2sinalphacosalpha=sin(2alpha)#

#[5]" "=color(blue)(1-sin(2x))#