Prove that $cos 3 x = 4 {cos}^{3} x - 3 cos x$ $cos 3x = 4cos^3 x - 3cos x$ ?

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Answer 1 · 2015-11-08T00:17:37

Prove $cos 3 x = 4 {cos}^{3} x - 3 cos x$ $cos 3x = 4cos^3 x - 3cos x$

Apply the trig identities:

$cos (a + b) = cos a \cdot cos b - sin a \cdot sin b$ $cos (a + b) = cos a * cos b - sin a * sin b$
$cos 2 x = 2 {cos}^{2} x - 1$ $cos 2x = 2cos^2 x - 1$
$sin 2 x = 2 sin x \cdot cos x$ $sin 2x = 2sin x * cos x$

We get:

$cos 3 x = cos (x + 2 x) = cos x \cdot cos 2 x - sin x \cdot sin 2 x$ $cos 3x = cos (x + 2x) = cos x * cos 2x - sin x * sin 2x$

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

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

$= 2 {cos}^{3} x - cos x - 2 cos x + 2 {cos}^{3} x$ $= 2cos^3 x - cos x - 2cos x + 2cos^3 x$

So

$cos 3 x = 4 {cos}^{3} x - 3 cos x$ $cos 3x = 4cos^3 x - 3cos x$

Prove that cos3x=4cos3x−3cosxcos 3x = 4cos^3 x - 3cos x ?