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Miscellaneous Exercise · Q4

Q.Prove that (cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=4sin⁡2x−y2(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\frac{x-y}{2}.

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Expand the left side using the Pythagorean identity, then apply the half-angle formula for sine to show both sides equal 2−2cos⁡(x−y)2 - 2\cos(x - y).

The heart of this proof lies in recognizing that the left side is a sum of squared differences. When we expand it, the cross terms will combine beautifully with the Pythagorean identity cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1. The right side, meanwhile, is begging us to use the half-angle formula. Once both sides are written in terms of cos⁡(x−y)\cos(x - y), the equality becomes transparent.

Proof

1. Expand the left-hand side

Start by squaring each binomial:

(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2(\cos x - \cos y)^2 + (\sin x - \sin y)^2

=cos⁡2x−2cos⁡xcos⁡y+cos⁡2y+sin⁡2x−2sin⁡xsin⁡y+sin⁡2y= \cos^2 x - 2\cos x \cos y + \cos^2 y + \sin^2 x - 2\sin x \sin y + \sin^2 y

2. Group the squared terms

Rearrange to collect the squares of xx and yy separately:

=(cos⁡2x+sin⁡2x)+(cos⁡2y+sin⁡2y)−2(cos⁡xcos⁡y+sin⁡xsin⁡y)= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y + \sin x \sin y)

3. Apply the Pythagorean identity

Since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 for any angle θ\theta:

=1+1−2(cos⁡xcos⁡y+sin⁡xsin⁡y)= 1 + 1 - 2(\cos x \cos y + \sin x \sin y)

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

4. Recognize the cosine difference formula

The expression cos⁡xcos⁡y+sin⁡xsin⁡y\cos x \cos y + \sin x \sin y is precisely cos⁡(x−y)\cos(x - y):

=2−2cos⁡(x−y)= 2 - 2\cos(x - y)

=2(1−cos⁡(x−y))= 2(1 - \cos(x - y)) …

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