Question

Verify the following trig identity: (equation in the comments)

Answer 1

\[\sin ^{3}A-\cos ^{3}A = (sinA-cosA)[1+\frac{ 1 }{ 2 }\sin(2A)]\]

Answer 2

factor first i guess

Answer 3

\[a^3-b^3=(a-b)(a^2+ab+b^2)\] is the first step

Answer 4

you get
\[\left(\sin(x)-\cos(x)\right)(\sin^2(x)+\sin(x)\cos(x)+\cos^2(x))\]

Answer 5

that give you half of what you want
then for the second factor use the fact that \(\sin^2(x)+\cos^2(x)=1\) to rewrite as
\[\left(\sin(x)-\cos(x)\right)(\sin^2(x)+\sin(x)\cos(x)+\cos^2(x))\]
\[=\left(\sin(x)-\cos(x)\right)(1+\sin(x)\cos(x))\]

Answer 6

double angle formula finishes it

Answer 7

Alright, got it. Thanks!