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Mathematics · Ch 3 — Trigonometry - II

Formulae for Conversion of Product into Sum or Difference

3.4.2

Formulae for Conversion of Product into Sum or Difference

Conversion of product into sum or difference

Theorem: for any angles AA and BB,

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)

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B=\cos(A-B)-\cos(A+B)

These are exactly the four identities used to derive the sum-to-product formulas in section 3.4.1, read in the

reverse direction: instead of starting from sin⁡(A+B)±sin⁡(A−B)\sin(A+B)\pm\sin(A-B) and simplifying to a product, we start from

a product 2sin⁡Acos⁡B2\sin A\cos B (etc.) and expand it into the sum/difference sin⁡(A+B)+sin⁡(A−B)\sin(A+B)+\sin(A-B) using the ordinary

compound-angle formulas of section 3.1, added or subtracted in pairs.

Solved Examples

Ex.2(i) — Express 2sin⁡4θcos⁡2θ2\sin4\theta\cos2\theta as a sum or difference. Direct substitution into

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B) with A=4θ,B=2θA=4\theta,B=2\theta: =sin⁡6θ+sin⁡2θ=\sin6\theta+\sin2\theta.

Ex.2(ii) — Express 4sin⁡A+B2sin⁡A−B24\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2} as a sum or difference. This is 2×(2sin⁡A+B2sin⁡A−B2)2\times\left(2\sin \dfrac{A+B}2\sin\dfrac{A-B}2\right); apply 2sin⁡Xsin⁡Y=cos⁡(X−Y)−cos⁡(X+Y)2\sin X\sin Y=\cos(X-Y)-\cos(X+Y) with X=A+B2,Y=A−B2X=\dfrac{A+B}2,Y=\dfrac{A-B}2:

X−Y=B, X+Y=AX-Y=B,\ X+Y=A, so the product is 2[cos⁡B−cos⁡A]=2cos⁡B−2cos⁡A2[\cos B-\cos A]=2\cos B-2\cos A. …

Misc Ex.2Express 2sin4θcos2θ and 4sin((A+B)/2)sin((A-B)/2) as a sum or difference

Worked out. First part is a direct one-line application of 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B=\sin(A+B)+\sin(A-B); second part applies 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B=\cos(A-B)-\cos(A+B) and simplifies the resulting angle arithmetic to 2(cos⁡B−cos⁡A)2(\cos B-\cos A). …