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Exercise 3.3 · Q16

Q.Prove that cos⁡9x−cos⁡5xsin⁡17x−sin⁡3x=−sin⁡2xcos⁡10x\dfrac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} = -\dfrac{\sin 2x}{\cos 10x}.

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Apply sum-to-product identities to both numerator and denominator, then simplify the resulting fractions to reveal the identity −sin⁡2xcos⁡10x-\dfrac{\sin 2x}{\cos 10x}.

The heart of this proof lies in recognizing that differences of cosines and sines can be rewritten as products using sum-to-product formulas. Once we transform both the numerator and denominator into product form, the expression simplifies beautifully through cancellation.

Sum-to-product identities convert sums or differences of trigonometric functions into products, which are far easier to manipulate algebraically. The key formulas we need are:

cos⁡A−cos⁡B=−2sin⁡(A+B2)sin⁡(A−B2)\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)

sin⁡A−sin⁡B=2cos⁡(A+B2)sin⁡(A−B2)\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)

Now let's work through the proof systematically.

Proof

  1. Transform the numerator using the cosine difference formula.

    For cos⁡9x−cos⁡5x\cos 9x - \cos 5x, we have A=9xA = 9x and B=5xB = 5x:

cos⁡9x−cos⁡5x=−2sin⁡(9x+5x2)sin⁡(9x−5x2)\cos 9x - \cos 5x = -2\sin\left(\frac{9x+5x}{2}\right)\sin\left(\frac{9x-5x}{2}\right)

=−2sin⁡(7x)sin⁡(2x)= -2\sin(7x)\sin(2x)

  1. Transform the denominator using the sine difference formula.

    For sin⁡17x−sin⁡3x\sin 17x - \sin 3x, we have A=17xA = 17x and B=3xB = 3x:

sin⁡17x−sin⁡3x=2cos⁡(17x+3x2)sin⁡(17x−3x2)\sin 17x - \sin 3x = 2\cos\left(\frac{17x+3x}{2}\right)\sin\left(\frac{17x-3x}{2}\right)

=2cos⁡(10x)sin⁡(7x)= 2\cos(10x)\sin(7x) …

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