Question 32 of 50intermediate🔧 ApplyLong Answer4 marks

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)
Correct Answer

L.H.S. =cot⁡4x (sin⁡5x+sin⁡3x)= \cot 4x \, (\sin 5x + \sin 3x)

Using the identity: sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right)

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

Therefore,

L.H.S.=cot⁡4x×2sin⁡4xcos⁡x\text{L.H.S.} = \cot 4x \times 2 \sin 4x \cos x =cos⁡4xsin⁡4x×2sin⁡4xcos⁡x= \dfrac{\cos 4x}{\sin 4x} \times 2 \sin 4x \cos x =2cos⁡4xcos⁡x= \mathbf{2 \cos 4x \cos x}

R.H.S. =cot⁡x (sin⁡5x−sin⁡3x)= \cot x \, (\sin 5x - \sin 3x)

Using the identity: sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right)

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

Therefore,

R.H.S.=cot⁡x×2cos⁡4xsin⁡x\text{R.H.S.} = \cot x \times 2 \cos 4x \sin x =cos⁡xsin⁡x×2cos⁡4xsin⁡x= \dfrac{\cos x}{\sin x} \times 2 \cos 4x \sin x =2cos⁡4xcos⁡x= \mathbf{2 \cos 4x \cos x}

Since L.H.S. = R.H.S., the identity is proved.

Exercise: EXERCISE 3.3 | Q: 15 | (Page 25)
For More Understanding

Explanation

This question tests the application of sum-to-product identities from the chapter. The key identities used are sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right) and sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2 \cos\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right), which are referenced in Example 16 of the textbook context.

The solution simplifies both sides independently to arrive at the same expression 2cos⁡4xcos⁡x2 \cos 4x \cos x, establishing the equality.

Solution Steps

  1. Step 1: Start with L.H.S. and apply sin⁡A+sin⁡B\sin A + \sin B identity to get 2sin⁡4xcos⁡x2 \sin 4x \cos x

  2. Step 2: Simplify using cot⁡4x=cos⁡4xsin⁡4x\cot 4x = \dfrac{\cos 4x}{\sin 4x} to get 2cos⁡4xcos⁡x2 \cos 4x \cos x

  3. Step 3: Take R.H.S. and apply sin⁡A−sin⁡B\sin A - \sin B identity to get 2cos⁡4xsin⁡x2 \cos 4x \sin x

  4. Step 4: Simplify using cot⁡x=cos⁡xsin⁡x\cot x = \dfrac{\cos x}{\sin x} to get 2cos⁡4xcos⁡x2 \cos 4x \cos x

  5. Step 5: Conclude L.H.S. = R.H.S. =2cos⁡4xcos⁡x= 2 \cos 4x \cos x, hence proved

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