How do you prove the identity $(sin^3x - cos^3x)/(sinx - cosx) =1 + sinxcosx$ ?

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Answer 1 · 2016-12-20T15:04:48

Start by factoring $sin^3x- cos^3x$ using the difference of cubes formula $a^3 - b^3 = (a- b)(a^2 + ab + b^2)$ .

So,

$(sin^3x - cos^3x)/(sinx - cosx) = 1 + sinxcosx$

$((sinx - cosx)(sin^2x + sinxcosx + cos^2x))/(sinx - cosx) = 1 + sinxcosx$

The $sinx - cosx$ cancel each other out.

$sin^2x + sinxcosx + cos^2x = 1 + sinxcosx$

Use the identity $sin^2theta + cos^2theta = 1$ :

$1 + sinxcosx = 1 + sinxcosx$

$LHS = RHS$

Hopefully this helps!

Answer 2 · 2016-12-20T15:06:02

See below.

From the polynomial identity

$(a^3-b^3)/(a-b)=a^2+a b + b^2$ we conclude that

$(sin^3x-cos^3x)/(sinx-cosx) = sin^2x+sin x cos x + cos^2x = 1+sinxcosx$

How do you prove the identity (sin^3x - cos^3x)/(sinx - cosx) =1 + sinxcosx(sin^3x - cos^3x)/(sinx - cosx) =1 + sinxcosx?