How do you verify (sin^3x - cos^3x) / (sinx - cosx) = (1 + sin2x) / 2?

1 Answer
Nov 2, 2015

Actually, that should be color(blue)((2+sin2x)/2) or color(red)(1+(sin2x)/2).

Explanation:

[1]" "(sin^3x-cos^3x)/(sinx-cosx)

Difference of two cubes: a^3-b^3=(a-b)(a^2+ab+b^2)

[2]" "=((sinx-cosx)(sin^2x+sinxcosx+cos^2x))/(sinx-cosx)

Cancel (sinx-cosx)

[3]" "=sin^2x+sinxcosx+cos^2x

Pythagorean Identity: sin^2theta+cos^2theta=1

[4]" "=1+sinxcosx

Double Angle Identity: sinthetacostheta=1/2sin2theta

[5]" "=1+1/2sin2x

Multiply 1 by 2/2 so the two terms will have the same denominator.

[6]" "=2/2+1/2sin2x

[7]" "=(2+sin2x)/2

color(magenta)(" ":. (sin^3x-cos^3x)/(sinx-cosx)=(2+sin2x)/2=1+(sin2x)/2)