Question

Prove that:
 $ \cot 4x\left( {\sin 5x + \sin 3x} \right) = \cot x\left( {\sin 5x - \sin 3x} \right) $

Answer 1

Hint: To prove required results we consider both sides of a given trigonometric equation. We first take left hand side of equation and then simplifying it by using trigonometric ratio and trigonometric identities and then we take right hand side and then simplifying it by using same concept and we that both sides on simplification gives out same result and hence we can say that left and right side of given equation are also equal.
Formulas used: $ \sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right),\,\,\sin C - \sin D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right) $

Complete step-by-step answer:
To prove the required result we will solve or simplify both sides of the given trigonometric equation.
Considering the left hand side of the given equation. We have,
 $ \cot 4x\left( {\sin 5x + \sin 3x} \right) $
Applying trigonometric identity on bracket part. We have,
 $
  \cot 4x\left[ {2\sin \left( {\dfrac{{5x + 3x}}{2}} \right)\cos \left( {\dfrac{{5x - 3x}}{2}} \right)} \right] \\
   \Rightarrow \cot 4x\left( {2\sin 4x\cos x} \right) \\
   \Rightarrow \dfrac{{\cos 4x}}{{\sin 4x}}\left( {2\sin 4x\cos x} \right) \\
   \Rightarrow \dfrac{{\cos 4x}}{{\sin 4x}} \times 2\sin 4x.\cos x \\
   \Rightarrow 2\cos 4x\cos x \\
  or\,\,we\,\,write \\
  \cot 4x\left( {\sin 5x + \sin 3x} \right) = 2\cos 4x\cos x............(i) \;
  $
Now, consider the right hand side of the given trigonometric equation. We have,
 $ \cot x\left( {\sin 5x - \sin 3x} \right) $
Applying trigonometric identity on bracket part. We have,
 $
  \cot x\left[ {2\cos \left( {\dfrac{{5x + 3x}}{2}} \right)\sin \left( {\dfrac{{5x - 3x}}{2}} \right)} \right] \\
   \Rightarrow \cot x\left( {2\cos 4x\sin x} \right) \\
   \Rightarrow \dfrac{{\cos x}}{{\sin x}}\left( {2\cos 4x\sin x} \right) \\
   \Rightarrow \dfrac{{\cos x}}{{\sin x}} \times 2\cos 4x\sin x \\
   \Rightarrow 2\cos x\cos 4x \\
  or\,\,we\,\,write \\
  \cot x\left( {\sin 5x - \sin 3x} \right) = 2\cos x\cos 4x............(ii) \;
  $
Therefore, from equation (i) and (ii) we see that the right hand side of both are the same.
Hence, their left hand side will also be equal.
So, we have
 $ \cot 4x\left( {\sin 5x + \sin 3x} \right) = \cot x\left( {\sin 5x - \sin 3x} \right) $
Hence, the required proof.

Note: For this type of problems in which angles of trigonometric terms are different on the left side as well as on the right hand side. So, we have to solve or simplify both sides by using identities and then from these equal results we can also say that the left and right side of a given equation is also equal.