Question

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

Answer 1

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)`