How do you express $sin 3 θ$ $sin 3theta$ in terms of trigonometric functions of $θ$ $theta$ ?

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How do you express $sin 3 θ$ $sin 3theta$ in terms of trigonometric functions of $θ$ $theta$ ?

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Answer 1 · 2018-01-04T20:30:52

$sin 3 x = 3 sin x {cos}^{2} x - {sin}^{3} x$ $sin3x=3sinxcos^2x-sin^3x$

Explanation:

.

$sin 3 x = sin (2 x + x) = sin 2 x cos x + cos 2 x sin x$ $sin3x=sin(2x+x)=sin2xcosx+cos2xsinx$

We have double-angle identities:

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$

$cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x=cos^2x-sin^2x$

$sin 3 x = 2 sin x cos x cos x + ({cos}^{2} x - {sin}^{2} x) sin x$ $sin3x=2sinxcosxcosx+(cos^2x-sin^2x)sinx$

$sin 3 x = 2 sin x {cos}^{2} x + sin x {cos}^{2} x - {sin}^{3} x$ $sin3x=2sinxcos^2x+sinxcos^2x-sin^3x$

$sin 3 x = 3 sin x {cos}^{2} x - {sin}^{3} x$ $sin3x=3sinxcos^2x-sin^3x$

Answer 2 · 2018-01-04T20:51:39

$sin 3 θ = 3 {cos}^{2} θ sin θ - {sin}^{3} θ = 3 sin θ - 4 {sin}^{3} θ$ $sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta = 3 sin theta - 4 sin^3 theta$

Explanation:

We can use de Moivre's formula:

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(cos theta + i sin theta)^n = cos n theta + i sin n theta$

as follows:

$cos 3 θ + i sin 3 θ$ $cos 3 theta + i sin 3 theta$

$= {(cos θ + i sin θ)}^{3}$ $= (cos theta + i sin theta)^3$

$= {cos}^{3} θ + 3 i {cos}^{2} θ sin θ - 3 cos θ {sin}^{2} θ - i {sin}^{3} θ$ $= cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin^2 theta - i sin^3 theta$

$= ({cos}^{3} θ - 3 cos θ {sin}^{2} θ) + i (3 {cos}^{2} θ sin θ - {sin}^{3} θ)$ $= (cos^3 theta - 3 cos theta sin^2 theta) + i (3 cos^2 theta sin theta - sin^3 theta)$

Then equating real and imaginary parts, we find:

${ (cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta), (sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta) :}$

We can also use the trigonometric version of Pythagoras' theorem:

$cos^2 theta + sin^2 theta = 1$

to refactor these formulae as follows:

$cos 3 theta = cos^3 theta - 3 cos theta sin^2 theta$

$color(white)(cos 3 theta) = cos^3 theta - 3 cos theta (1-cos^2 theta)$

$color(white)(cos 3 theta) = 4cos^3 theta - 3 cos theta$

$sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta$

$color(white)(sin 3 theta) = 3 (1-sin^2 theta) sin theta - sin^3 theta$

$color(white)(sin 3 theta) = 3 sin theta - 4 sin^3 theta$

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How do you express $sin 3 θ$ $sin 3theta$ in terms of trigonometric functions of $θ$ $theta$ ?

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