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## Answers

now sin^2x+cos^2x=1

also

2cosA*cosB=cos(A+B)+cos(A-B)

2sinA*sinB=cos(A-b)-cos(A+B)

so

(sin^2α+cos^2α)+(sin^2β+cos^2β)+cos(α+β)+cos(α-β)+cos(α-β)-cos(α+β)

2+2cos(α-β)

now

cos2x=2cos^2x-1

sooooooooooo

2+2(2cos^2(α-β)/2-1)

4cos^2(α-β)/2