Question 31 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

sin⁡2x+2sin⁡4x+sin⁡6x=4cos⁡2xsin⁡4x\sin 2x + 2 \sin 4x + \sin 6x = 4 \cos^2 x \sin 4x
Correct Answer

L.H.S. =sin⁡2x+2sin⁡4x+sin⁡6x= \sin 2x + 2 \sin 4x + \sin 6x

=(sin⁡2x+sin⁡6x)+2sin⁡4x= (\sin 2x + \sin 6x) + 2 \sin 4x

Using the identity sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2):

=2sin⁡(2x+6x2)cos⁡(6x−2x2)+2sin⁡4x= 2 \sin\left(\frac{2x+6x}{2}\right) \cos\left(\frac{6x-2x}{2}\right) + 2 \sin 4x =2sin⁡4xcos⁡2x+2sin⁡4x= 2 \sin 4x \cos 2x + 2 \sin 4x =2sin⁡4x(cos⁡2x+1)= 2 \sin 4x (\cos 2x + 1)

From the identity cos 2x = 2cos² x - 1:

cos⁡2x+1=2cos⁡2x−1+1=2cos⁡2x\cos 2x + 1 = 2\cos^2 x - 1 + 1 = 2\cos^2 x

Therefore,

L.H.S.=2sin⁡4x(2cos⁡2x)\text{L.H.S.} = 2 \sin 4x (2\cos^2 x) =4cos⁡2xsin⁡4x=R.H.S.= \mathbf{4 \cos^2 x \sin 4x} = \text{R.H.S.}

Hence proved.

Exercise: EXERCISE 3.3 | Q: 14 | (Page 25)
For More Understanding

Explanation

This question tests the application of sum-to-product identities and double angle formulas. The key strategy is to group sin⁡2x\sin 2x and sin⁡6x\sin 6x together, apply the sin⁡A+sin⁡B\sin A + \sin B identity to create a common factor of sin⁡4x\sin 4x, then use the double angle formula for cosine to simplify cos⁡2x+1\cos 2x + 1 into 2cos⁡2x2\cos^2 x.

This approach is demonstrated in Example 15 and Example 16 from the textbook context.

Solution Steps

  1. Step 1: Group sin⁡2x+sin⁡6x\sin 2x + \sin 6x and apply sum-to-product identity sin⁡A+sin⁡B=2sin⁡(A+B2)cos⁡(A−B2)\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)

  2. Step 2: Simplify to get 2sin⁡4xcos⁡2x+2sin⁡4x2 \sin 4x \cos 2x + 2 \sin 4x

  3. Step 3: Factor out 2sin⁡4x2 \sin 4x to get 2sin⁡4x(cos⁡2x+1)2 \sin 4x (\cos 2x + 1)

  4. Step 4: Use identity cos⁡2x=2cos⁡2x−1\cos 2x = 2\cos^2 x - 1 to simplify cos⁡2x+1=2cos⁡2x\cos 2x + 1 = 2\cos^2 x

  5. Step 5: Multiply to get final result 4cos⁡2xsin⁡4x4 \cos^2 x \sin 4x

PreviousNext