Skip to content
Exercise 3.3 · Q15

Q.Prove that cot⁡4x (sin⁡5x+sin⁡3x)=cot⁡x (sin⁡5x−sin⁡3x)\cot 4x\,(\sin 5x + \sin 3x) = \cot x\,(\sin 5x - \sin 3x).

Uttar Pradesh UpmspTextbookSubjective· 3mImportance★★★★★est
27% · 41/150 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Use sum-to-product formulas to rewrite both sides in terms of products of sines and cosines, then simplify using cot⁡=cos⁡sin⁡\cot = \frac{\cos}{\sin} to show both sides equal 2cos⁡4xcos⁡x2\cos 4x \cos x.

We need to prove an identity involving cotangents and sums/differences of sines. The natural instinct is to convert the sums of sines into products — that’s the classic trick when you see sin⁡A±sin⁡B\sin A \pm \sin B. Once we do that, the cot⁡\cot terms will combine nicely, and we can check if both sides simplify to the same expression.

Let’s go step by step.

  1. Left-hand side (LHS): cot⁡4x (sin⁡5x+sin⁡3x)\cot 4x\,(\sin 5x + \sin 3x) Use the sum-to-product formula:

sin⁡5x+sin⁡3x=2sin⁡(5x+3x2)cos⁡(5x−3x2)=2sin⁡4xcos⁡x\sin 5x + \sin 3x = 2 \sin\left(\frac{5x+3x}{2}\right) \cos\left(\frac{5x-3x}{2}\right) = 2 \sin 4x \cos x

So LHS becomes:

cot⁡4x⋅(2sin⁡4xcos⁡x)=cos⁡4xsin⁡4x⋅2sin⁡4xcos⁡x=2cos⁡4xcos⁡x\cot 4x \cdot (2 \sin 4x \cos x) = \frac{\cos 4x}{\sin 4x} \cdot 2 \sin 4x \cos x = 2 \cos 4x \cos x

  1. Right-hand side (RHS): cot⁡x (sin⁡5x−sin⁡3x)\cot x\,(\sin 5x - \sin 3x) Use the difference-to-product formula:

sin⁡5x−sin⁡3x=2cos⁡(5x+3x2)sin⁡(5x−3x2)=2cos⁡4xsin⁡x\sin 5x - \sin 3x = 2 \cos\left(\frac{5x+3x}{2}\right) \sin\left(\frac{5x-3x}{2}\right) = 2 \cos 4x \sin x

So RHS becomes:

cot⁡x⋅(2cos⁡4xsin⁡x)=cos⁡xsin⁡x⋅2cos⁡4xsin⁡x=2cos⁡4xcos⁡x\cot x \cdot (2 \cos 4x \sin x) = \frac{\cos x}{\sin x} \cdot 2 \cos 4x \sin x = 2 \cos 4x \cos x

  1. Comparison: Both sides simplify to exactly 2cos⁡4xcos⁡x2 \cos 4x \cos x. Hence the identity holds for all xx where the original expressions are defined (i.e., where sin⁡4x≠0\sin 4x \neq 0 and sin⁡x≠0\sin x \neq 0). …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.