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Exercise: Sum and Difference Formulas · Q17

Q.Prove that sin⁡(x+y)sin⁡(x−y)=sin⁡2x−sin⁡2y\sin(x+y)\sin(x-y) = \sin^2x - \sin^2y.

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Expand each factor:

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

a difference of squares. Replace cos⁡2y=1−sin⁡2y\cos^2y=1-\sin^2y and cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x:

=sin⁡2x(1−sin⁡2y)−(1−sin⁡2x)sin⁡2y=sin⁡2x−sin⁡2xsin⁡2y−sin⁡2y+sin⁡2xsin⁡2y=sin⁡2x−sin⁡2y.= \sin^2x(1-\sin^2y) - (1-\sin^2x)\sin^2y = \sin^2x - \sin^2x\sin^2y - \sin^2y + \sin^2x\sin^2y = \sin^2x-\sin^2y.

✓Final answer

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

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