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Exercise 3.3 · Q20

Q.Prove that sin⁡x−sin⁡3xsin⁡2x−cos⁡2x=2sin⁡x\dfrac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x} = 2\sin x.

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The key is to rewrite the numerator using the sine difference identity and the denominator using sin⁡2x−cos⁡2x=−(cos⁡2x−sin⁡2x)=−cos⁡2x\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x. After simplification, the expression reduces to 2sin⁡x2\sin x.


When you see a trigonometric expression like this, the first instinct should be: can I rewrite everything in terms of a common angle or a common function? Here, the numerator has sin⁡x\sin x and sin⁡3x\sin 3x — that’s a difference of sines with different angles. The denominator has sin⁡2x−cos⁡2x\sin^2 x - \cos^2 x, which is a classic form that screams “double-angle identity.”

The trick is to simplify both parts separately, then cancel.


  1. Simplify the denominator first. Recall the double-angle identity: cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x. Our denominator is sin⁡2x−cos⁡2x\sin^2 x - \cos^2 x, which is exactly the negative of that:

sin⁡2x−cos⁡2x=−(cos⁡2x−sin⁡2x)=−cos⁡2x.\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x.

So the denominator becomes −cos⁡2x-\cos 2x.

  1. Now simplify the numerator. We have sin⁡x−sin⁡3x\sin x - \sin 3x. Use the sine difference 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}.

Here A=xA = x, B=3xB = 3x, so:

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

Since sin⁡(−x)=−sin⁡x\sin(-x) = -\sin x, this becomes:

sin⁡x−sin⁡3x=2cos⁡2x⋅(−sin⁡x)=−2cos⁡2xsin⁡x.\sin x - \sin 3x = 2 \cos 2x \cdot (-\sin x) = -2 \cos 2x \sin x. …

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