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Prove the triple angle identities for sine and cosine

by RoRi

Prove the triple angle identities for the sine and cosine:

    \[ \sin 3x = 3 \sin x - 4 \sin^3 x \qquad \text{and} \qquad \cos 3x = \cos x - 4 \sin^2 x \cos x \quad \text{for all } x \in \mathbb{R}. \]

Also, prove the following alternative version of the triple angle identity for the cosine,

    \[ \cos 3x = 4 \cos^3 x - 3 \cos x. \]

Proof. First, the triple angle identity for the sine function.

    \begin{align*} \sin 3x = \sin (2x + x) &= \cos x \sin 2x + \sin x \cos 2x & (\text{Thm 2.3 (f)})\\ &= \cos x (2 \sin x \cos x) + \sin x (1 - 2\sin^2 x) &(\text{Double angle formulas})\\ &= 2 \sin x \cos^2 x + \sin x - 2 \sin^3 x \\ &= 2 \sin x (1 - \sin^2 x) + \sin x - 2 \sin^3 x \\ &= 2 \sin x - 2 \sin^3 x + \sin x - 2 \sin^3 x \\ &= 3 \sin x - 4 \sin^3 x. \end{align*}

Next, the first triple angle identity for the cosine function.

    \begin{align*} \cos 3x = \cos(2x + x) &= \cos 2x \cos x - \sin 2x \sin x & (\text{Thm 2.3 (f)})\\ &= (1- 2 \sin^2 x)\cos x - 2 \sin^2 x \cos x & (\text{Double angle formulas})\\ &= \cos x - 2 \sin^2 x \cos x - 2 \sin^2 x \cos x \\ &= \cos x - 4 \sin^2 x \cos x. \end{align*}

Finally, the alternative version of the triple angle identity for the cosine function. We start with the version we have above and apply the pythagorean identity, \sin^2 + \cos^2 = 1 \implies \sin^2 = 1 - \cos^2,

    \begin{align*} \cos 3x = \cos x - 4 \sin^2 x \cos x &= \cos x - 4(1-\cos^2 x) \cos x \\ &= \cos x - 4 \cos x + 4 \cos^3 x \\ &= 4 \cos^3 x - 3 \cos x. \qquad \blacksquare \end{align*}

One comment

  1. Anonymous says:

    Easier Proof: Use Euler’s formula by saying

    e^(3ix) = (e^(ix))^3
    cos(3x) + isin(3x) = [cos(x) + isin(x)]^3

    Expand the cube and equate the complex part with the complex to get sin(3x) and the real part with real part to get cos(3x).

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