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Worked Examples · Example 17

Q.Prove that sin⁡5x−2sin⁡3x+sin⁡xcos⁡5x−cos⁡x=tan⁡x\dfrac{\sin 5x - 2\sin 3x + \sin x}{\cos 5x - \cos x} = \tan x.

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The expression simplifies to tan⁡x\tan x by first rewriting the numerator as 2sin⁡xcos⁡2x−2sin⁡3x2\sin x \cos 2x - 2\sin 3x and the denominator as −2sin⁡3xsin⁡2x-2\sin 3x \sin 2x, then factoring and cancelling common terms.

This is a classic trigonometric identity proof where the trick lies in recognising that the numerator and denominator can be expressed in terms of sum-to-product formulas. The key insight: instead of expanding sin⁡5x\sin 5x and sin⁡3x\sin 3x individually (which gets messy), we group terms to apply identities that convert sums into products.

Let’s work through it step by step.

  1. Focus on the numerator first: sin⁡5x−2sin⁡3x+sin⁡x\sin 5x - 2\sin 3x + \sin x

    Group the first and last terms: (sin⁡5x+sin⁡x)−2sin⁡3x(\sin 5x + \sin x) - 2\sin 3x

    Use the sum-to-product identity: sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

    Here A=5xA=5x, B=xB=x, so A+B2=3x\frac{A+B}{2}=3x and A−B2=2x\frac{A-B}{2}=2x

    Thus sin⁡5x+sin⁡x=2sin⁡3xcos⁡2x\sin 5x + \sin x = 2\sin 3x \cos 2x

    The numerator becomes 2sin⁡3xcos⁡2x−2sin⁡3x=2sin⁡3x(cos⁡2x−1)2\sin 3x \cos 2x - 2\sin 3x = 2\sin 3x (\cos 2x - 1)

  2. Simplify cos⁡2x−1\cos 2x - 1: Recall the identity cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x, so cos⁡2x−1=−2sin⁡2x\cos 2x - 1 = -2\sin^2 x

    Therefore numerator = 2sin⁡3x⋅(−2sin⁡2x)=−4sin⁡3xsin⁡2x2\sin 3x \cdot (-2\sin^2 x) = -4\sin 3x \sin^2 x

  3. Now the denominator: cos⁡5x−cos⁡x\cos 5x - \cos x

    Use the sum-to-product identity: cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

    Here A=5xA=5x, B=xB=x, so A+B2=3x\frac{A+B}{2}=3x and A−B2=2x\frac{A-B}{2}=2x

    Thus cos⁡5x−cos⁡x=−2sin⁡3xsin⁡2x\cos 5x - \cos x = -2\sin 3x \sin 2x

  4. Form the fraction:

−4sin⁡3xsin⁡2x−2sin⁡3xsin⁡2x\frac{-4\sin 3x \sin^2 x}{-2\sin 3x \sin 2x}

The negatives cancel, and sin⁡3x\sin 3x cancels (provided sin⁡3x≠0\sin 3x \neq 0, but the identity holds wherever both sides are defined). …

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