Question 45 of 50advanced🔧 ApplyLong Answer4 marks

Prove that:

sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x=4cos⁡xcos⁡2xsin⁡4x\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \cos x \cos 2x \sin 4x
Correct Answer

L.H.S. =sin⁡x+sin⁡3x+sin⁡5x+sin⁡7x= \sin x + \sin 3x + \sin 5x + \sin 7x

We group the terms strategically:

=(sin⁡x+sin⁡7x)+(sin⁡3x+sin⁡5x)= (\sin x + \sin 7x) + (\sin 3x + \sin 5x)

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

For sin⁡x+sin⁡7x\sin x + \sin 7x:

=2sin⁡(x+7x2)cos⁡(x−7x2)= 2 \sin\left(\frac{x + 7x}{2}\right) \cos\left(\frac{x - 7x}{2}\right) =2sin⁡(4x)cos⁡(−3x)= 2 \sin(4x) \cos(-3x) =2sin⁡(4x)cos⁡(3x)= 2 \sin(4x) \cos(3x)

For sin⁡3x+sin⁡5x\sin 3x + \sin 5x:

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

Therefore,

L.H.S.=2sin⁡(4x)cos⁡(3x)+2sin⁡(4x)cos⁡(x)\text{L.H.S.} = 2 \sin(4x) \cos(3x) + 2 \sin(4x) \cos(x) =2sin⁡(4x)[cos⁡(3x)+cos⁡(x)]= 2 \sin(4x) \left[\cos(3x) + \cos(x)\right]

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

cos⁡(3x)+cos⁡(x)=2cos⁡(2x)cos⁡(x)\cos(3x) + \cos(x) = 2 \cos(2x) \cos(x)

Therefore,

L.H.S.=2sin⁡(4x)×2cos⁡(2x)cos⁡(x)\text{L.H.S.} = 2 \sin(4x) \times 2 \cos(2x) \cos(x) =4cos⁡xcos⁡2xsin⁡4x= 4 \cos x \cos 2x \sin 4x =R.H.S.= \textbf{R.H.S.}

Hence proved.

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 5 | (Page 29)
For More Understanding

Explanation

This question requires applying 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(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) and cos⁡A+cos⁡B=2cos⁡(A+B2)cos⁡(A−B2)\cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right), which are demonstrated in Examples 15 and 16 of the textbook context.

The strategy is to group the four sine terms into two pairs, apply the identity to each pair, factor out the common term sin⁡(4x)\sin(4x), and then apply the cosine sum-to-product identity to reach the RHS.

Solution Steps

  1. Step 1: Group terms as (sin⁡x+sin⁡7x)+(sin⁡3x+sin⁡5x)(\sin x + \sin 7x) + (\sin 3x + \sin 5x)

  2. Step 2: Apply sin⁡A+sin⁡B\sin A + \sin B identity to each pair to get 2sin⁡(4x)cos⁡(3x)+2sin⁡(4x)cos⁡(x)2 \sin(4x) \cos(3x) + 2 \sin(4x) \cos(x)

  3. Step 3: Factor out 2sin⁡(4x)2 \sin(4x) to get 2sin⁡(4x)[cos⁡(3x)+cos⁡(x)]2 \sin(4x)\left[\cos(3x) + \cos(x)\right]

  4. Step 4: Apply cos⁡A+cos⁡B\cos A + \cos B identity to get 2sin⁡(4x)×2cos⁡(2x)cos⁡(x)2 \sin(4x) \times 2 \cos(2x) \cos(x)

  5. Step 5: Simplify to get 4cos⁡xcos⁡2xsin⁡4x=4 \cos x \cos 2x \sin 4x = R.H.S.

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