Express $\sin 3\theta$ and $\cos 3\theta$ as functions of $\sin \theta$ and $\cos \theta$ using Euler's identity

Question

Using Euler's identity ($e^{in\theta}=\cos n\theta+i \sin n\theta$), express $\sin 3\theta$ and $\cos 3\theta$ as functions of $\sin \theta$ and $\cos \theta$.

Any ideas?

Answer 1

HINT:

So, $$\cos3x+i\sin3x=e^{i3x}=(e^{ix})^3=(\cos x+i\sin x)^3$$

Use Binomial Expansion and equate the real & the imaginary parts

Answer 2

$e^{in\theta}=\cos n\theta+i \sin n\theta$, therefore $e^{-in\theta}=\cos -n\theta+i \sin -n\theta=\cos n\theta-i \sin n\theta$. Therefore, $\cos n\theta=\frac{e^{in\theta}+e^{-n\theta}}{2}$ and $\sin n\theta=\frac{e^{in\theta}-e^{-n\theta}}{i2}$

From Euler's identity, $\cos 3\theta+i \sin 3\theta=e^{i3\theta}=(e^{i\theta})^3=(\cos \theta+i \sin \theta)^3$, therefore $\cos 3\theta=(\cos \theta+i \sin \theta)^3-isin 3\theta=(\cos \theta+i \sin \theta)^3-\frac{(\cos \theta+i \sin \theta)^3-(\cos \theta-i \sin \theta)^3}{2}$.

Similar, $\sin 3\theta=(\cos \theta+i \sin \theta)^3-cos 3\theta=(\cos \theta+i \sin \theta)^3-\frac{(\cos \theta+i \sin \theta)^3+(\cos \theta-i \sin \theta)^3}{2}$.