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Miscellaneous Exercise · Q2

Q.Prove that (sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡x=0(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0.

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✓ Free question

The identity simplifies to zero by expanding the products, then using sum-to-product formulas or directly applying triple-angle identities to cancel every term — the final result is 00.

The key insight here is that you don't need to memorise a trick. Each term looks like a product of a sum/difference of trig functions with a single function. The natural instinct is to expand, then simplify using known identities for sin⁡3x\sin 3x and cos⁡3x\cos 3x in terms of sin⁡x\sin x and cos⁡x\cos x. Alternatively, you can use sum-to-product formulas to combine the brackets first — both paths lead to the same clean cancellation.

Let’s work through the expansion method first, since it’s the most direct.

  1. Expand the expression Write E=(sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡xE = (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x. Expanding:

E=sin⁡3xsin⁡x+sin⁡2x+cos⁡3xcos⁡x−cos⁡2x.E = \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x.

  1. Group the sin⁡3xsin⁡x\sin 3x \sin x and cos⁡3xcos⁡x\cos 3x \cos x terms Notice that sin⁡3xsin⁡x+cos⁡3xcos⁡x\sin 3x \sin x + \cos 3x \cos x looks like the cosine of a difference:

cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B.\cos(A - B) = \cos A \cos B + \sin A \sin B.

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

sin⁡3xsin⁡x+cos⁡3xcos⁡x=cos⁡(3x−x)=cos⁡2x.\sin 3x \sin x + \cos 3x \cos x = \cos(3x - x) = \cos 2x.

Tip

Recognising the cosine difference identity here saves you from expanding sin⁡3x\sin 3x and cos⁡3x\cos 3x fully — a neat shortcut.

So now:

E=cos⁡2x+sin⁡2x−cos⁡2x.E = \cos 2x + \sin^2 x - \cos^2 x.

  1. Simplify sin⁡2x−cos⁡2x\sin^2 x - \cos^2 x

    Recall the double-angle identity: cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = \cos^2 x - \sin^2 x.

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

    Substitute:

E=cos⁡2x+(−cos⁡2x)=0.E = \cos 2x + (-\cos 2x) = 0.

  1. Conclusion The expression simplifies exactly to zero, independent of xx (except where undefined, but it’s defined for all real xx).
Watch out

A common mistake is to misapply the sign in the cosine difference identity. Remember: cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B) = \cos A \cos B + \sin A \sin B — the plus sign is correct. If you accidentally use a minus, you’ll get cos⁡4x\cos 4x instead of cos⁡2x\cos 2x, and the cancellation won’t happen.

✓Final answer

The expression (sin⁡3x+sin⁡x)sin⁡x+(cos⁡3x−cos⁡x)cos⁡x(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x equals 00 for all xx.

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