Question #9c7b3

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Question #9c7b3

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Answer 1 · 2016-11-25T11:49:24

${sin}^{3} x = \frac{3 sin x - sin (3 x)}{4}$ $sin^3x=(3sinx-sin(3x))/4$

Explanation:

If you know the de Moivre's identity it is quite direct.

$e^{i x} = cos x + i sin x$ $e^(i x) = cosx+isinx$

so

$e^{3 i x} = {(cos x + i sin x)}^{3} = cos (3 x) + i sin (3 x)$ $e^(3ix)= (cosx + isinx)^3=cos(3x)+isin(3x)$

or

${cos}^{3} x + 3 i {cos}^{2} x sin x - 3 cos x {sin}^{2} x - i {sin}^{3} x = cos (3 x) + i sin (3 x)$ $cos^3x+3icos^2xsinx-3cosxsin^2x-isin^3x=cos(3x)+isin(3x)$

grouping the real and the imaginary parts,

${cos}^{3} x - 3 cos x {sin}^{2} x = cos (3 x)$ $cos^3x-3cosxsin^2x=cos(3x)$ and
$3 {cos}^{2} x sin x - {sin}^{3} x = sin (3 x)$ $3cos^2xsinx-sin^3x=sin(3x)$

now, using the identity

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$ we will get at

$sin (3 x) = 3 sin x (1 - {sin}^{2} x) - {sin}^{3} x$ $sin(3x)=3sinx(1-sin^2x)-sin^3x$ or

${sin}^{3} x = \frac{3 sin x - sin (3 x)}{4}$ $sin^3x=(3sinx-sin(3x))/4$

Answer 2 · 2016-11-25T14:32:17

Expand ${sin}^{3} x$ $sin^3 x$

Explanation:

There is another simpler method.
Use the trig identity:
$sin 3 x = 3 sin x - 4 {sin}^{3} x$ $sin 3x = 3sin x - 4sin^3 x$
By transposing terms, we get:
$4 {sin}^{3} x = 3 sin x - sin 3 x$ $4sin^3 x = 3sin x - sin 3x$
${sin}^{3} x = \frac{3 sin x - sin 3 x}{4}$ $sin^3 x = (3sin x - sin 3x)/4$

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