Question #1b893

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Answer 1 · 2017-02-27T21:35:06

see below

Use the following formulas:

Left Hand Side:

$cos 3 θ = cos (2 θ + θ)$ $cos 3 theta = cos (2theta+theta)$

$= cos 2 θ cos θ - sin 2 θ sin θ$ $=cos2theta cos theta - sin 2 theta sin theta$

$= (2 {cos}^{2} θ - 1) cos θ - 2 sin θ cos θ \cdot sin θ$ $=(2cos^2theta-1)cos theta-2sin theta cos theta*sin theta$

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

$= 2 {cos}^{3} θ - cos θ - 2 cos θ (1 - {cos}^{2} θ)$ $=2cos^3 theta-cos theta-2cos theta (1-cos^2 theta)$

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

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

$:.=$ Right Hand Side

Answer 2 · 2017-02-27T21:53:45

See below.

De Moivre's Formula:

$(cos x +i sin x )^n =\cos nx +i\sin nx$

So:

$(cos x +i sin x )^3 =\cos 3x +i\sin 3x qquad square$

But it is also true from a Binomial Expansion that:

$(cos x +i sin x )^3 =cos^3 x + 3 cos^2 x (i sin x) + 3 cos x (i sin x)^2 + (i sin x)^3$

$=cos^3 x + 3 i cos^2 x sin x - 3 cos x sin^2 x - i sin^3 x$

Collecting real and imaginary terms:

$=(cos^3 x - 3 cos x sin^2 x)+ i \ (3 cos^2 x sin x - sin^3 x) qquad triangle$

Equating real terms in $triangle$ and $square$ :

$implies cos 3x = cos^3 x - 3 cos x sin^2 x$

Using the Pythagorean Identity:

$implies cos 3x = cos^3 x - 3 cos x color(red)((1 - cos^2 x))$

$implies cos 3x = cos^3 x - 3 cos x + 3cos^3 x$

$implies cos 3x = 4 cos^3 x - 3 cos x$

Ta da!!

Of course, replace $x$ with $theta$