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Exercise 3.3 · Q17

Q.Prove that sin⁡5x+sin⁡3xcos⁡5x+cos⁡3x=tan⁡4x\dfrac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x.

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The identity sin⁡5x+sin⁡3xcos⁡5x+cos⁡3x=tan⁡4x\dfrac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x is proved by applying the sum-to-product formulas for sine and cosine, which simplify the numerator and denominator into a single ratio that directly gives tan⁡4x\tan 4x.

This is a classic trigonometric identity proof. The key insight is that when you have sums of sines or cosines with different angles, the sum-to-product formulas are your best friend. They let you rewrite the sum as a product, and in this case, the common factor cancels beautifully, leaving a simple tangent.

Let’s work through it step by step.

  1. Identify the pattern.

    The numerator is sin⁡5x+sin⁡3x\sin 5x + \sin 3x and the denominator is cos⁡5x+cos⁡3x\cos 5x + \cos 3x. Both are sums of two trigonometric functions with angles 5x5x and 3x3x. The sum-to-product formulas are designed exactly for this situation.

  2. Apply the sum-to-product formula for sine.

    The formula is:

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)

Here, A=5xA = 5x and B=3xB = 3x. So:

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

  1. Apply the sum-to-product formula for cosine. The formula is:

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)

Again, with A=5xA = 5x and B=3xB = 3x:

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

  1. Form the ratio. Now substitute these into the original expression:

sin⁡5x+sin⁡3xcos⁡5x+cos⁡3x=2sin⁡(4x)cos⁡(x)2cos⁡(4x)cos⁡(x)\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}

  1. Cancel the common factors. The factor 22 cancels, and cos⁡(x)\cos(x) cancels (provided cos⁡x≠0\cos x \neq 0, but the identity holds wherever both sides are defined). This leaves:

sin⁡(4x)cos⁡(4x)\frac{\sin(4x)}{\cos(4x)}

  1. Recognize the definition of tangent. …

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