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Worked Examples · Example 16

Q.Prove that cos⁡7x+cos⁡5xsin⁡7x−sin⁡5x=cot⁡x\dfrac{\cos 7x + \cos 5x}{\sin 7x - \sin 5x} = \cot x.

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Use sum-to-product identities to rewrite the numerator and denominator as products, then simplify the ratio to cot⁡x\cot x.

The key to this proof is recognising that both the numerator and denominator are sums or differences of trigonometric functions of different angles. When you see expressions like cos⁡A+cos⁡B\cos A + \cos B or sin⁡A−sin⁡B\sin A - \sin B, the standard trick is to convert them into products using the sum-to-product formulas. That turns a messy ratio into something much simpler — often a single trigonometric ratio.

Let’s apply that idea step by step.

  1. Apply sum-to-product to the numerator. The formula for the sum of two cosines is:

cos⁡P+cos⁡Q=2cos⁡P+Q2cos⁡P−Q2\cos P + \cos Q = 2 \cos\frac{P+Q}{2} \cos\frac{P-Q}{2}

Here P=7xP = 7x and Q=5xQ = 5x, so:

cos⁡7x+cos⁡5x=2cos⁡7x+5x2cos⁡7x−5x2=2cos⁡6xcos⁡x\cos 7x + \cos 5x = 2 \cos\frac{7x+5x}{2} \cos\frac{7x-5x}{2} = 2 \cos 6x \cos x

  1. Apply sum-to-product to the denominator. The formula for the difference of two sines is:

sin⁡P−sin⁡Q=2cos⁡P+Q2sin⁡P−Q2\sin P - \sin Q = 2 \cos\frac{P+Q}{2} \sin\frac{P-Q}{2}

With P=7xP = 7x and Q=5xQ = 5x:

sin⁡7x−sin⁡5x=2cos⁡7x+5x2sin⁡7x−5x2=2cos⁡6xsin⁡x\sin 7x - \sin 5x = 2 \cos\frac{7x+5x}{2} \sin\frac{7x-5x}{2} = 2 \cos 6x \sin x …

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