Question

Prove that $\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$ .

Answer 1

Hint: For solving this question, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side. And we will use trigonometric formulas of $\sin C+\sin D$ , $\cos C-\cos D$ and $\cos 2\theta $ for simplifying the term on the left-hand side. After that, we will easily prove the desired result.

Complete step-by-step answer:
Given:
We have to prove the following equation:
$\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)..................\left( 1 \right) \\
 & \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...............\left( 2 \right) \\
 & \cos 2\theta =1-2{{\sin }^{2}}\theta ......................................................\left( 3 \right) \\
 & \sin 2\theta =2\sin \theta \cos \theta .....................................................\left( 4 \right) \\
 & \dfrac{\sin \theta }{\cos \theta }=\tan \theta ................................................................\left( 5 \right) \\
\end{align}$
Now, we will use the above five formulas to simplify the term on the left-hand side.
On the left-hand side we have $\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}$ . Then,
$\begin{align}
  & \dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x} \\
 & \Rightarrow \dfrac{\sin 5x+\sin x-2\sin 3x}{\cos 5x-\cos x} \\
\end{align}$
Now, we will use the formula from the equation (1) to write $\sin 5x+\sin x=2\sin 3x\cos 2x$ and formula from the equation (2) to write $\cos 5x-\cos x=-2\sin 3x\sin 2x$ in the above expression. Then,
$\begin{align}
  & \dfrac{\sin 5x+\sin x-2\sin 3x}{\cos 5x-\cos x} \\
 & \Rightarrow \dfrac{2\sin \left( \dfrac{5x+x}{2} \right)\cos \left( \dfrac{5x-x}{2} \right)-2\sin 3x}{-2\sin \left( \dfrac{5x+x}{2} \right)\sin \left( \dfrac{5x-x}{2} \right)} \\
 & \Rightarrow \dfrac{2\sin 3x\cos 2x-2\sin 3x}{-2\sin 3x\sin 2x} \\
\end{align}$
Now, we can take $-2\sin 3x$ common from each term in the numerator of the above expression. Then,
$\begin{align}
  & \dfrac{2\sin 3x\cos 2x-2\sin 3x}{-2\sin 3x\sin 2x} \\
 & \Rightarrow \dfrac{-2\sin 3x\left( 1-\cos 2x \right)}{-2\sin 3x\sin 2x} \\
 & \Rightarrow \dfrac{1-\cos 2x}{\sin 2x} \\
\end{align}$
Now, we will use the formula from the equation (3) to write $1-\cos 2x=2{{\sin }^{2}}x$ and $\sin 2x=2\sin x\cos x$ in the above expression. Then,
$\begin{align}
  & \dfrac{1-\cos 2x}{\sin 2x} \\
 & \Rightarrow \dfrac{2{{\sin }^{2}}x}{2\sin x\cos x} \\
 & \Rightarrow \dfrac{\sin x}{\cos x} \\
\end{align}$
Now, we will use the formula from the equation (5) to write $\dfrac{\sin x}{\cos x}=\tan x$ in the above expression. Then,
$\begin{align}
  & \dfrac{\sin x}{\cos x} \\
 & \Rightarrow \tan x \\
\end{align}$
Now, from the above result we conclude that the value of the expression $\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}$ will be equal to the value of the expression $\tan x$ . Then,
$\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$
Now, from the above result we conclude that, the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\dfrac{\sin 5x-2\sin 3x+\sin x}{\cos 5x-\cos x}=\tan x$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like $\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)$ correctly. Moreover, while simplifying we should be aware of the result and avoid calculation mistakes while solving.