Estimate value of $\sin\left(\frac{\theta}{3}\right)$ using taylor polynomial

Question

Question: We may write $[\cos(\theta)+ i\sin(\theta)]^n = cos^n(x)(1+i\tan(\theta))^n$
Use the 5th order Taylor polynomial for $(1+x)^\frac{1}{3}$ with $x = i\tan(\theta)$ to find an estimate for $\sin\left(\frac{\theta}{3}\right)$.
Use this approximation to estimate the value of $\sin\left(\frac{\pi}{18}\right)$.
Hint: Two complex numbers $a+bi$ and $c+di$ are equal if $a=c$ and $b=d$.

This is what I've com up with... $T_5(1+i\tan\theta)^{1/3} = [1 - \frac{\tan^2(\theta)}{9} - \frac{10}{243}(\tan^4\theta)] + i[\frac{\tan\theta}{3} -\frac{5}{81}\tan^3\theta+ (\frac{22}{729})(\tan^5\theta)] $

$\sin(\frac{\theta}{3}) =[\frac{\sin\theta}{3} -\frac{5}{81}\sin^3\theta+ (\frac{22}{729})(\sin^5\theta)] $

$sin(\frac{\pi}{18}) =\left[\frac{\sin\frac{\pi}{6}}{3} -\frac{5}{81}\sin^3\frac{\pi}{6}+ (\frac{22}{729})(\sin^5\frac{\pi}{6})\right] $

Is this correct or am I missing something?

Answer 1

Your Taylor approximation is good, but you've misapplied it. From the formula that you've given: $$\sin\left(\frac{\theta}{3}\right) \approx \cos^{1/3}(\theta)\left(\frac{1}{3}\tan(\theta)-\frac{5}{81}\tan^3(\theta)+\frac{22}{729}\tan^5(\theta)\right)$$ This is because $[\cos(\theta)+i\sin(\theta)]^n = \cos(n\theta)+i\sin(n\theta)$. Note that the $\cos^{1/3}(\theta)$ does not cancel out with the tangent terms to produce sines. Then, $$\sin\left(\frac{\pi}{18}\right) \approx \cos^{1/3}\left(\frac{\pi}{6}\right)\left(\frac{1}{3}\tan\left(\frac{\pi}{6}\right)-\frac{5}{81}\tan^3\left(\frac{\pi}{6}\right)+\frac{22}{729}\tan^5\left(\frac{\pi}{6}\right)\right)$$ Using the fact that $\cos(\pi/6) = \frac{\sqrt{3}}{2}$ and that $\tan(\pi/6) = \frac{1}{3\sqrt{3}}$, we can (tediously) find that the right-hand side is exactly $\frac{1037}{6561}\sqrt{\frac{4}{3}}\approx 0.173962$. This is pretty good, as $\sin\left(\frac{\pi}{18}\right) \approx 0.173648$.