# Triple Angle Identities

The trigonometric **triple-angle identities** give a relationship between the basic trigonometric functions applied to three times an angle in terms of trigonometric functions of the angle itself.

Triple-angle Identities\[ \sin 3 \theta = 3 \sin \theta - 4 \sin ^3 \theta \] \[ \cos 3\theta = 4 \cos ^ 3 \theta - 3 \cos \theta \]

To prove the triple-angle identities, we can write \(\sin 3 \theta\) as \(\sin(2 \theta + \theta)\). Then we can use the sum formula and the double-angle identities to get the desired form:

\[\begin{align} \sin 3 \theta & = \sin (2 \theta + \theta)\\ & = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta \\ & = (2 \sin \theta \cos \theta) \cos \theta + \big(1 - 2 \sin^2 \theta\big) \sin \theta \\ & = 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta \big(1 - \sin^2 \theta\big) + \sin \theta - 2 \sin^3 \theta \\ & = 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - 2 \sin^3 \theta \\ & = 3 \sin \theta - 4 \sin^3 \theta.\ _\square \end{align} \]

The triple angle identity of \(\cos 3 \theta\) can be proved in a very similar manner.

From these formulas, we also have the following identities for \(\sin^3(\theta)\) and \(\cos^3 (\theta) \) in terms of lower powers:

\[\sin^3(\theta) = \frac{3 \sin (\theta) - \sin \left( 3 \theta \right) }{4},\quad \cos^3 (\theta) = \frac{\cos(3\theta) + 3 \cos (\theta)}{4}.\]

To remember the cosine formula, the trick that I like to use is to read cosine as "dollar." Then, we say

"Dollar thirty is equal to four dollar thirty minus three dollar."

\[ \begin{array} {l l l l l } $1. 30 & = $ 4.30 & - $ 3 \\ 1 \cos 3 \theta & = 4 \cos ^3 \theta & - 3 \cos \theta \\ \end{array} \]

Just leaving a mark here,

\[ \begin{eqnarray} \tan x \tan(60^\circ - x) \tan (60^\circ + x) &=& \tan(3x) \\ 4\sin x \sin(60^\circ - x) \sin (60^\circ + x) &=&\sin(3x) \\ 4\cos x \cos(60^\circ - x) \cos(60^\circ + x) &=& \cos(3x) \\ \tan(3x) &=& \frac {3\tan x - \tan^3 x}{1 - 3\tan^2 x}. \end{eqnarray} \]

Prove that

\[ \tan (6^\circ) = \tan (12^\circ) \tan(24^\circ) \tan(48^\circ ). \]

**Cite as:**Triple Angle Identities.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/triple-angle-identities/