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# Show that $\dfrac{\sin x}{cos3x}+\dfrac{\sin 3x}{cos9x}+\dfrac{\sin 9x}{cos27x}=\dfrac{1}{2}\left[ \tan 27x-\tan x \right]$ ?

Last updated date: 26th Feb 2024
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Hint: These types of problems are pretty straight forward and are very simple to solve. Problems like these require prior knowledge of trigonometric equations and other various trigonometric formulae. The main catch of the problem is to multiply the numerator and denominator of each and every term in the left hand side by a factor and then to proceed further with it and convert it then into the required tangent form by multiplying and dividing $2$ on the numerator and denominator.

Now, we start off the solution to the given problem as,
We consider the first term of the left hand side, $\dfrac{\sin x}{cos3x}$ . We multiply and divide the numerator and the denominator with $\cos x$ , it then transforms to,
$\dfrac{\sin x\cos x}{cos3x\cos x}$ . Now, multiplying and dividing the numerator and denominator by $2$ we get,
\begin{align} & \dfrac{1}{2}\dfrac{2\sin x\cos x}{cos3x\cos x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 2x}{cos3x\cos x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 3x-x \right)}{cos3x\cos x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 3x\cos x-\cos 3x\sin x}{cos3x\cos x} \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 3x\cos x}{cos3x\cos x}-\dfrac{\cos 3x\sin x}{cos3x\cos x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 3x}{cos3x}-\dfrac{\sin x}{\cos x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \tan 3x-\tan x \right] \\ \end{align}
Now, simplifying the second term by following the same steps we get,
\begin{align} & \dfrac{1}{2}\dfrac{2\sin 3x\cos 3x}{cos9x\cos 3x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 6x}{cos9x\cos 3x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 9x-3x \right)}{cos9x\cos 3x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 9x\cos 3x-\cos 9x\sin 3x}{cos9x\cos 3x} \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 9x\cos 3x}{cos9x\cos 3x}-\dfrac{\cos 9x\sin 3x}{cos9x\cos 3x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 9x}{cos9x}-\dfrac{\sin 3x}{\cos 3x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \tan 9x-\tan 3x \right] \\ \end{align}
Now, simplifying the third term by following the same steps we get,
\begin{align} & \dfrac{1}{2}\dfrac{2\sin 9x\cos 9x}{cos27x\cos 9x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 18x}{cos27x\cos 9x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin \left( 27x-9x \right)}{cos27x\cos 9x} \\ & \Rightarrow \dfrac{1}{2}\dfrac{\sin 27x\cos 9x-\cos 27x\sin 9x}{cos27x\cos 9x} \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 27x\cos 9x}{cos27x\cos 9x}-\dfrac{\cos 27x\sin 9x}{cos27x\cos 9x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \dfrac{\sin 27x}{cos27x}-\dfrac{\sin 9x}{\cos 9x} \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \tan 27x-\tan 9x \right] \\ \end{align}
Now adding all the three terms in their simplified form we get,
$\dfrac{1}{2}\left[ \tan 3x-\tan x \right]+\dfrac{1}{2}\left[ \tan 9x-\tan 3x \right]+\dfrac{1}{2}\left[ \tan 27x-\tan 9x \right]$
Adding up all these we get,
\begin{align} & \Rightarrow \dfrac{1}{2}\left[ \tan 3x-\tan x+\tan 9x-\tan 3x+\tan 27x-\tan 9x \right] \\ & \Rightarrow \dfrac{1}{2}\left[ \tan 27x-\tan x \right] \\ \end{align}
Thus, it gets evaluated to the above value. Now if we take a closer look, we see that the equation that we have arrived at is the same as that written in the right hand side of the given problem. Thus we say that we have proved the required equation correctly.

Note: For problems like these, we need to recall all the formulae and equations of trigonometry. The main catch of this problem is the multiplication of a term in both the numerator and denominator to simplify our problem. This idea of multiplication of a common term, will come only by more and more practice, or else it would be very difficult for us to solve these types of problems.