How do you simplify #cos^4alpha+2cos^2alphasin^2alpha+sin^4alpha#?

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Answer 1 · 2016-12-08T14:15:43

1

consider

#(x+y)^2=x^2+2xy+y^2#

let#" " x=cos^2alpha" "y=sin^2alpha " "# on RHS

#cos^4alpha+2cos^2alphasin^2alpha+sin^4alpha#

we have
#=(cos^2alpha+sin^2alpha)^2=1^2=1#