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

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Answer 1 · 2015-11-02T10:27:59

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

#[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)#