Question

Prove that:
$\dfrac{\sin x}{\cos 3x}+\dfrac{\sin 3x}{\cos 9x}+\dfrac{\sin 9x}{\cos 27x}=\dfrac{1}{2}\left[ \tan 27x-\tan x \right]$

Answer 1

First of all, multiply and divide the L.H.S of the above equation to 2. And the multiply and divide $\cos x$ with $\dfrac{\sin x}{\cos 3x}$, $\cos 3x$ with $\dfrac{\sin 3x}{\cos 9x}$ and $\cos 9x$ with $\dfrac{\sin 9x}{\cos 27x}$. Then we are going to use the trigonometric property which states that $\sin 2x=2\sin x\cos x$. Also, we need this trigonometric identity which states that $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.

Complete step-by-step solution:
The equation we are asked to prove is as follows:
$\dfrac{\sin x}{\cos 3x}+\dfrac{\sin 3x}{\cos 9x}+\dfrac{\sin 9x}{\cos 27x}=\dfrac{1}{2}\left[ \tan 27x-\tan x \right]$
We are going to rearrange the L.H.S of the above equation in such a way so that it will be equal to R.H.S. For that, we are going to multiply and divide 2 to L.H.S of the above equation and we get,
$\dfrac{2}{2}\left( \dfrac{\sin x}{\cos 3x}+\dfrac{\sin 3x}{\cos 9x}+\dfrac{\sin 9x}{\cos 27x} \right)$
Now, moving the 2 written in the numerator inside the bracket we get,
$\dfrac{1}{2}\left( \dfrac{2\sin x}{\cos 3x}+\dfrac{2\sin 3x}{\cos 9x}+\dfrac{2\sin 9x}{\cos 27x} \right)$
Now, multiplying and dividing $\cos x$ with $\dfrac{\sin x}{\cos 3x}$, $\cos 3x$ with $\dfrac{\sin 3x}{\cos 9x}$ and $\cos 9x$ with $\dfrac{\sin 9x}{\cos 27x}$ in the above expression and we get,
$\dfrac{1}{2}\left( \dfrac{2\sin x\cos x}{\cos 3x\cos x}+\dfrac{2\sin 3x\cos 3x}{\cos 9x\cos 3x}+\dfrac{2\sin 9x\cos 9x}{\cos 27x\cos 9x} \right)$
We know the trigonometry double angle property which states that:
$\sin 2x=2\sin x\cos x$
Using the above relation in the above expression we get,
$\dfrac{1}{2}\left( \dfrac{\sin 2x}{\cos 3x\cos x}+\dfrac{\sin 6x}{\cos 9x\cos 3x}+\dfrac{\sin 18x}{\cos 27x\cos 9x} \right)$
We are also going to use the sine identity which states that:
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
Writing $\sin 2x=\sin \left( 3x-x \right),\sin 6x=\sin \left( 9x-3x \right),\sin 18x=\sin \left( 27x-9x \right)$ in the above expression we get,
$\begin{aligned} & \dfrac{1}{2}\left( \dfrac{\sin \left( 3x-x \right)}{\cos 3x\cos x}+\dfrac{\sin \left( 9x-3x \right)}{\cos 9x\cos 3x}+\dfrac{\sin \left( 27x-9x \right)}{\cos 27x\cos 9x} \right) \\\ & =\dfrac{1}{2}\left( \dfrac{\sin 3x\cos x-\cos 3x\sin x}{\cos 3x\cos x}+\dfrac{\sin 9x\cos 3x-\cos 9x\sin 3x}{\cos 9x\cos 3x}+\dfrac{\sin 27x\cos 9x-\cos 27x\sin 9x}{\cos 27x\cos 9x} \right) \\\ \end{aligned}$
Rearranging the above expression and we get,
$=\dfrac{1}{2}\left( \tan 3x-\tan x+\tan 9x-\tan 3x+\tan 27x-\tan 9x \right)$
Terms with opposite sign get canceled out and we get,
$=\dfrac{1}{2}\left( \tan 27x-\tan x \right)$
As you can see that our L.H.S is coming equal to R.H.S so we have proved the given equation.

Note: To solve the above problem, we must know the following trigonometric identities:
$\sin 2x=2\sin x\cos x$
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
You cannot move forward in the above problem if you don’t know the above properties so make sure you have properly understood these concepts.