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Exercise 3.4 · Q65

Q.Prove the following: sin⁡2x−sin⁡2yx=tan⁡(x+y)tan⁡(x−y)\dfrac{\sin^2x-\sin^2y}{\phantom{x}}=\tan(x+y)\tan(x-y)

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Concept understanding — Transformation of Sum or Difference of Trigonometric Functions into Product and Vice Versa

Factorization (also called transformation) formulae convert a SUM or DIFFERENCE of two sines or cosines into a PRODUCT, and, conversely, convert a PRODUCT of two sines/cosines into a sum or difference. Substituting A=C+D2A=\dfrac{C+D}{2}, B=C−D2B=\dfrac{C-D}{2} into the four compound-angle identities gives sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C+\sin D=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}, sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}, cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C+\cos D=2\cos\dfrac{C+D}{2}\cos\dfrac{C-D}{2} and cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}. Reading the same four identities the other way around gives the product-to-sum forms 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B), 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B), 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos(A+B)+\cos(A-B) and $2\sin A\sin B=\cos(A-B)-\cos( …

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