# Find an expression for cos 3x in terms of cosx ?

Apr 6, 2017

It can be rewritten in terms of two addition identities:

$\sin \left(u + v\right) = \sin u \cos v + \cos u \sin v$
$\cos \left(u + v\right) = \cos u \cos v - \sin u \sin v$

$\sin \left(3 x\right) = \sin \left(2 x + x\right)$

$= \sin 2 x \cos x + \cos 2 x \sin x$

From the identities above, we have:

$\sin \left(2 x\right) = 2 \sin x \cos x$
$\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x$

Hence, we have:

$\sin \left(3 x\right) = \left(2 \sin x \cos x\right) \cos x + \left({\cos}^{2} x - {\sin}^{2} x\right) \sin x$

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

$= 3 \sin x {\cos}^{2} x - {\sin}^{3} x$

$= 3 \sin x \left(1 - {\sin}^{2} x\right) - {\sin}^{3} x$

$= \textcolor{b l u e}{3 \sin x - 4 {\sin}^{3} x}$

Apr 6, 2017

$\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$

#### Explanation:

Another approach to the excellent answer from @Truong-Son N. , which is less work should you need higher powers/multiples is to use de Moivre's theorem, using complex numbers.

de Moivre's theorem states that for any complex number $z$ in trigonometric form $z = \cos \theta + i \sin \theta$, then

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

With $n = 3$ we have:

${\left(\cos \theta + i \sin \theta\right)}^{3} = \cos 3 \theta + i \sin 3 \theta$

And expanding the LHS using the binomial theorem we have:

${\left(\cos \theta\right)}^{3} + 3 {\left(\cos \theta\right)}^{2} \left(i \sin \theta\right) + 3 \left(\cos \theta\right) {\left(i \sin \theta\right)}^{2} + {\left(i \sin \theta\right)}^{3} = \cos 3 \theta + i \sin 3 \theta$

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

$\therefore \left({\cos}^{3} \theta - 3 \cos \theta {\sin}^{2} \theta\right) + i \left(3 {\cos}^{2} \theta \sin \theta - {\sin}^{3} \theta\right) = \cos 3 \theta + i \sin 3 \theta$

If we equate imaginary components we get:

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

As we want the expression in terms of $\sin \theta$ we can replace ${\cos}^{2} \theta$ using the fundamental identity ${\sin}^{2} A + {\cos}^{2} A \equiv 1$ to get:

$\sin 3 \theta = 3 \left(1 - {\sin}^{2} \theta\right) \sin \theta - {\sin}^{3} \theta$
$\text{ } = 3 \sin \theta - 3 {\sin}^{3} \theta - {\sin}^{3} \theta$
$\text{ } = 3 \sin \theta - 4 {\sin}^{3} \theta$

Or,

$\sin 3 x = 3 \sin x - 4 {\sin}^{3} x$

Incidentally, as an extension we also get an expression for $\cos 3 x$ for free! Equating real components we get:

$\cos 3 \theta = {\cos}^{3} \theta - 3 \cos \theta {\sin}^{2} \theta$
$\text{ } = {\cos}^{3} \theta - 3 \cos \theta \left(1 - {\cos}^{2} \theta\right)$
$\text{ } = {\cos}^{3} \theta - 3 \cos \theta + 3 {\cos}^{3} \theta$
$\text{ } = 4 {\cos}^{3} \theta - 3 \cos \theta$

Or,

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