Question 47 of 50advanced🔧 ApplyLong Answer4 marks

Prove that:

sin⁡3x+sin⁡2x−sin⁡x=4sin⁡xcos⁡x2cos⁡3x2\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}
Correct Answer

We need to prove that sin⁡3x+sin⁡2x−sin⁡x=4sin⁡xcos⁡x2cos⁡3x2\sin 3x + \sin 2x - \sin x = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2}.

LHS =sin⁡3x+sin⁡2x−sin⁡x= \sin 3x + \sin 2x - \sin x.

Group terms: (sin⁡3x−sin⁡x)+sin⁡2x(\sin 3x - \sin x) + \sin 2x.

Using the identity sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}, we get:

sin⁡3x−sin⁡x=2cos⁡3x+x2sin⁡3x−x2=2cos⁡(2x)sin⁡x\sin 3x - \sin x = 2 \cos \frac{3x+x}{2} \sin \frac{3x-x}{2} = 2 \cos(2x) \sin x

Also, from the given formulas, sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x.

Thus,

LHS=2cos⁡(2x)sin⁡x+2sin⁡xcos⁡x=2sin⁡x(cos⁡2x+cos⁡x)\text{LHS} = 2 \cos(2x) \sin x + 2 \sin x \cos x = 2 \sin x (\cos 2x + \cos x)

Now apply the identity cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}:

cos⁡2x+cos⁡x=2cos⁡2x+x2cos⁡2x−x2=2cos⁡3x2cos⁡x2\cos 2x + \cos x = 2 \cos \frac{2x+x}{2} \cos \frac{2x-x}{2} = 2 \cos \frac{3x}{2} \cos \frac{x}{2}

Substituting back:

LHS=2sin⁡x⋅2cos⁡3x2cos⁡x2=4sin⁡xcos⁡x2cos⁡3x2=RHS\text{LHS} = 2 \sin x \cdot 2 \cos \frac{3x}{2} \cos \frac{x}{2} = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2} = \textbf{RHS}

Hence proved.

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 7 | (Page 29)
For More Understanding

Explanation

The proof uses standard sum-to-product identities for sine and cosine, which are part of the NCERT curriculum and are referenced in the provided context (e.g., Example 15 uses Identity 20(i) for cosine sum).

The step-by-step manipulation groups terms to apply these identities efficiently, reducing the expression to the desired form.

Solution Steps

  1. Step 1: Write LHS =sin⁡3x+sin⁡2x−sin⁡x=(sin⁡3x−sin⁡x)+sin⁡2x= \sin 3x + \sin 2x - \sin x = (\sin 3x - \sin x) + \sin 2x.

  2. Step 2: Apply sin⁡A−sin⁡B\sin A - \sin B identity: sin⁡3x−sin⁡x=2cos⁡(2x)sin⁡x\sin 3x - \sin x = 2 \cos(2x) \sin x.

  3. Step 3: Use sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x from the given formulas.

  4. Step 4: Combine: LHS=2sin⁡x(cos⁡2x+cos⁡x)\text{LHS} = 2 \sin x (\cos 2x + \cos x).

  5. Step 5: Apply cos⁡A+cos⁡B\cos A + \cos B identity: cos⁡2x+cos⁡x=2cos⁡3x2cos⁡x2\cos 2x + \cos x = 2 \cos \frac{3x}{2} \cos \frac{x}{2}.

  6. Step 6: Multiply: LHS=4sin⁡xcos⁡x2cos⁡3x2=RHS\text{LHS} = 4 \sin x \cos \frac{x}{2} \cos \frac{3x}{2} = \textbf{RHS}.

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