Find the value of $\frac{\sin 24x}{\cos18x \cos 6x}$.

Question

Given $\tan x = \sqrt{5} - 2 $, find the value of $$\frac{\sin 24x}{\cos 18x\cos 6x}.$$

Now I wrote numerator as $\sin (18 + 6)x$ and I used identity $\sin(A+B)$, after that I reduced problem to $\tan 18x + \tan 6x$

How do I proceed? Thanks

Answer 1

We have

\begin{align} \tan 2x&=\frac{2\tan x}{1-\tan^2x}\\ &=\frac{2(\sqrt{5}-2)}{1-(\sqrt{5}-2)^2}\\ &=\frac{2(\sqrt{5}-2)}{1-5+4\sqrt{5}-4}\\ &=\frac{2(\sqrt{5}-2)}{4(\sqrt{5}-2)}\\ &=\frac{1}{2} \end{align}

Then use

$$\tan 3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}$$

to deduce that

$$\tan 6x=\frac{11}{2}$$

Use it once again to deduce that

$$\tan18x=\frac{1199}{718}$$

Answer 2

$$\tan2x=\frac{2(\sqrt5-2)}{1-(\sqrt5-2)^2}=\frac{1}{2},$$ which gives $$\sin4x=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}.$$

Answer 3

hint

$$\tan (2x)=2\frac{\sqrt {5}-2}{4\sqrt {5}-8}=\frac {1}{2} $$

Answer 4

HINT:

Let us try to rationalize

$$(\tan x+2)^2=5$$

$$\iff\tan^2x-4\tan x-1=0\iff1=2\cdot\dfrac{2\tan x}{1-\tan^2x}=2\tan2x$$