Trig identity question (1 Viewer)

highshill · Jul 13, 2017

Does any one how to solve this I know what do but keep making a mistake somewhere
Prove:
(sins-cosx)= sinx-cosx-2sinxcosx+2sinxcos^2x

1729 · Jul 13, 2017

highshill said:
Does any one how to solve this I know what do but keep making a mistake somewhere
Prove:
(sins-cosx)= sinx-cosx-2sinxcosx+2sinxcos^2x

Double-check the LHS is written correctly ('sins', and there may be an exponent because there are parentheses)

highshill · Jul 14, 2017

It was the LHS to the power of three sorry I typed it wrong

He-Mann · Jul 14, 2017

Identity is false for all integer powers (of LHS).

He-Mann · Jul 14, 2017

I believe there is another typo on the RHS. Expand via binomial theorem and use Pythagorean trig identities to get

$\begin{align*} (\sin x - \cos x)^3 &= \sin^3 x - 3 \sin^2 x \cos x + 3 \sin x \cos^2 x - \cos^3 x \\ &= \sin x (1 - \cos^2 x) - 3 \sin^2 x \cos x + 3 \sin x \cos^2 x - \cos x (1 - \sin^2 x) \\ &= \sin x - \cos x - 2 \sin^2 x \cos x + 2 \sin x \cos^2 x\end{align*}$

Trig identity question (1 Viewer)

highshill

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1729

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highshill

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He-Mann

Vexed?

He-Mann

Vexed?

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