# Trigonometrycos^3 x

#### dwsmith

##### Well-known member
How does one go about writing $\cos^3 x$ as a linear combination of $\cos x$ and $\cos 3x\mbox{?}$

#### SuperSonic4

##### Well-known member
MHB Math Helper
How does one go about writing $\cos^3 x$ as a linear combination of $\cos x$ and $\cos 3x\mbox{?}$
The triple angle identity for cos(3x) is $\cos(3x) = 4\cos^3(x) - 3\cos(x)$. From there it's algebraic manipulation (unless I misunderstood the question).

If you want/have to verify that identity for yourself note that $\cos(3x) = \cos(2x+x)$ and use the double angle/addition and Pythagorean identities where appropriate.

edit: tex of course

#### Prove It

##### Well-known member
MHB Math Helper
An alternative method: Using DeMoivre's Theorem...

\displaystyle \begin{align*} \left( \cos{\theta} + i\sin{\theta} \right)^n &= \cos{n\theta} + i\sin{n\theta} \\ \left( \cos{\theta} + i\sin{\theta} \right)^3 &= \cos{3\theta} + i\sin{3\theta} \\ \cos^3{\theta} + 3i\cos^2{\theta}\sin{\theta} + 3i^2 \cos{\theta}\sin^2{\theta} + i^3\sin^3{\theta} &= \cos{3\theta} + i\sin{3\theta} \\ \cos^3{\theta} - 3\cos{\theta}\sin^2{\theta} + i\left( 3\cos^2{\theta}\sin{\theta} - \sin^3{\theta} \right) &= \cos{3\theta} + i\sin{3\theta} \\ \cos^3{\theta} - 3\cos{\theta}\left( 1 - \cos^2{\theta} \right) + i\left[ 3\left( 1 - \sin^2{\theta} \right) \sin{\theta} - \sin^3{\theta} \right] &= \cos{3\theta} + i\sin{3\theta} \\ 4\cos^3{\theta} - 3\cos{\theta} + i\left( 3\sin{\theta} - 4\sin^3{\theta} \right) &= \cos{3\theta} + i\sin{3\theta} \end{align*}

When we equate real and imaginary parts we get $\displaystyle \cos{3\theta} = 4\cos^3{\theta} - 3\cos{\theta}$ and $\displaystyle \sin{3\theta} = 3\sin{\theta} - 4\sin^3{\theta}$.