Trignometric problem (using De Movier's Theorem)

Question

Ok so this question, I started out writing tan as sin and cos in the right side of the equation, simplified as much as possible and ended up with a very (sort of) fascinating equation which is

Where s = sin thetha and c= sin thetha

As you can see the denominator and numerator looks very simple where on top sin x is with the power 5 and in the bottom cos x is with the power 5.

Then I went onto applying the De Movire theorem

And then ended up with-

As you can see the required terms sin 5 thetha and cos 5 thetha is there, but now all I need is to cancel out those two other terms in both denominator and numerator. I simply cant find a way in which those cancel off, My question is-

How do I proceed from here?? And if I had done something wrong (if those dont cancel) then where did I go wrong? Please help

Answer 1

Using DeMoivre's Theorem and the Binomial Theorem, you get:

$\cos 5\theta + i\sin 5\theta = (\cos \theta + i\sin \theta)^5$

$= \cos^5\theta + 5i\cos^4\theta\sin\theta + 10i^2\cos^3 \theta\sin^2\theta + 10i^3\cos^2 \theta\sin^3\theta + 5i^4\cos \theta \sin^4 \theta + i^5\sin^5\theta$

$= \cos^5\theta + 5i\cos^4\theta\sin \theta - 10\cos^3 \theta\sin^2\theta - 10i\cos^2 \theta\sin^3\theta + 5\cos\theta\sin^4 \theta + i\sin^5\theta$

$= (\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4 \theta)+i(5\cos^4\theta\sin \theta - 10\cos^2 \theta\sin^3\theta + \sin^5\theta)$

Equate real and imaginary parts to get expressions for $\cos 5\theta$ and $\sin 5\theta$. Can you finish from here?

Answer 2

This may be easier if you start with the De Moivre equation

$$\cos (5\theta) + i \sin(5\theta) = (\cos \theta + i \sin \theta)^5$$

For simplicity, let $c = \cos \theta$ and $s = \sin \theta$. Then, the RHS is simply $(c + is)^5$. Expanding it (using the binomial coefficients and that $i^2 = -1$) gives:

$$c^5+5 i c^4 s-10 c^3 s^2-10 i c^2 s^3+5 c s^4+i s^5$$

or

$$(c^5 - 10 c^3 s^2 - 5cs^4) + i (5c^4s-10c^2s^3+s^5)$$

Equating the real and imaginary parts of the original equation gives

$$\cos (5\theta) = c^5 - 10 c^3 s^2 - 5cs^4$$ $$\sin (5\theta) = 5c^4s-10c^2s^3+s^5$$

Now, what is $\tan (5\theta)$ in terms of the above?