Can someone double check that the question in (a) (ii) is wrong, because I got
$\cos 3 \theta = 4 \cos^3 \theta -3 \cos \theta $ after I compared the real parts, and I got $\sin 3 \theta = 3 \sin \theta -4 \sin^3 \theta $ after comparing imaginary parts.
However, I need to check if the given (a)(ii) is right or is my working right, as I am stuck on (b) and (c).
Using my results, for (b), I tried doing
$\frac{\sin3\theta-\sin\theta}{\cos3\theta+\cos\theta}$=$\frac{2 \sin\theta-4\sin ^3\theta}{4\cos^3\theta-2\cos\theta}$ which wouldn't get me anywhere near $tan \theta$. Please advise.
And would anyone be able to suggest how I attempt part (c) as well?
Also, sorry in advance for any wrong title or tags (trying to improve on it!)
Best Answer
Yes, it's obvious that the formula is miswritten. If you set $\theta = \pi/6$ you'll see:
$$\cos 3\theta = \cos \pi/2 = 0 \ne 4 \cos 3\theta - 3\cos \theta = 4\cos \pi/2 - 3\cos \pi/6 = -3\cos\pi/6$$
It look's like the printer had problems with superscripts, I too get the same result as you. So where they intend $\cos^3\theta$ they printed $\cos 3\theta$. Unfortunately the printing error results in ambiguity as they sometimes actually means $\cos 3\theta$.
The (b) depends on the corrected formulas from (a) and second seems OK (here they actually means $\sin 3\theta$ and $\cos 3\theta$):
$${2\sin \theta - 4\sin^3 \theta\over 4\cos^3\theta - 2\cos\theta} = {\sin\theta\over\cos\theta}{1-2\sin^2\theta\over 2\cos^2\theta-1} = {1-2\sin^2\theta\over2(1-\sin^2\theta)-1}\tan \theta = {1-2\sin^2\theta\over1-2\sin^2\theta}\tan \theta$$
For (c) you'll use the formulas formulas you found in (a) for $\sin 3\theta$ and $\cos 3\theta$ and insert in $\tan 3\theta = {\sin3\theta \over \cos 3\theta}$.