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Sum and Difference Formulas and Product-to-Sum Formulas of Trigonometric Functions 📂Functions

Sum and Difference Formulas and Product-to-Sum Formulas of Trigonometric Functions

The sum-to-product and product-to-sum formulas aren’t used as often as the double angle/half angle formulas, so they’re not considered as important. However, this doesn’t mean they’re entirely unnecessary. Since the derivation process is very simple, it’s good to be familiar with it and be able to derive it quickly whenever needed. They are derived using only the addition formulas.

Addition Formulas

sin⁡(θ1±θ2)= sin⁡θ1cos⁡θ2±sin⁡θ2cos⁡θ2cos⁡(θ1±θ2)= cos⁡θ1cos⁡θ2∓sin⁡θ1sin⁡θ2tan⁡(θ1±θ2)= tan⁡θ1±tan⁡θ21∓tan⁡θ1tan⁡θ2 \begin{align*} \sin ( \theta_{1} \pm \theta_{2}) =&\ \sin \theta_{1} \cos \theta_{2} \pm \sin \theta_{2} \cos \theta_{2} \\ \cos ( \theta_{1} \pm \theta_{2}) =&\ \cos \theta_{1} \cos\theta_{2} \mp \sin\theta_{1} \sin\theta_{2} \\ \tan ( \theta_{1} \pm \theta_{2}) =&\ \dfrac{\tan\theta_{1} \pm \tan\theta_{2}}{1 \mp \tan\theta_{1}\tan\theta_{2}} \end{align*}

Sum-to-Product Identities

sin⁡A+sin⁡B= 2sin⁡A+B2cos⁡A−B2sin⁡A−sin⁡B= 2cos⁡A+B2sin⁡A−B2cos⁡A+cos⁡B= 2cos⁡A+B2cos⁡A−B2cos⁡A−cos⁡B= −2sin⁡A+B2sin⁡A−B2 \begin{align*} \sin A + \sin B =&\ 2 \sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2} \\ \sin A - \sin B =&\ 2 \cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2} \\ \cos A + \cos B =&\ 2 \cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2} \\ \cos A - \cos B =&\ -2 \sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2} \end{align*}

Derivation

It’s lengthy but not difficult.

sin⁡(A+B)= sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)= sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡(A+B)= cos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡(A−B)= cos⁡Acos⁡B+sin⁡Asin⁡B \begin{align} \sin (A+B) =&\ \sin A \cos B + \cos A \sin B \\ \sin (A-B) =&\ \sin A \cos B - \cos A \sin B \\ \cos (A+B) =&\ \cos A \cos B - \sin A \sin B \\ \cos (A-B) =&\ \cos A \cos B + \sin A \sin B \end{align}

By substituting A≡x+y2A \equiv \dfrac{x+y}{2}, B≡x−y2B \equiv \dfrac{x-y}{2}, and applying (1)(1) to (4)(4), we get the following.

sin⁡x= sin⁡x+y2cos⁡x−y2+cos⁡x+y2sin⁡x−y2sin⁡y= sin⁡x+y2cos⁡x−y2−cos⁡x+y2sin⁡x−y2cos⁡x= cos⁡x+y2cos⁡x−y2−sin⁡x+y2sin⁡x−y2cos⁡y= cos⁡x+y2cos⁡x−y2+sin⁡x+y2sin⁡x−y2 \begin{align} \sin x =&\ \sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2} + \cos \dfrac{x+y}{2} \sin \dfrac{x-y}{2} \\ \sin y =&\ \sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2} - \cos \dfrac{x+y}{2} \sin \dfrac{x-y}{2} \\ \cos x =&\ \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2} - \sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2} \\ \cos y =&\ \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2} + \sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2} \end{align}

Calculating (5)+(6)(5)+(6) gives us the equation below.

sin⁡x+sin⁡y=2sin⁡x+y2cos⁡x−y2 \sin x + \sin y = 2\sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}

Calculating (5)−(6)(5)-(6) gives us the equation below.

sin⁡x+sin⁡y=2cos⁡x+y2sin⁡x−y2 \sin x + \sin y = 2\cos \dfrac{x+y}{2} \sin \dfrac{x-y}{2}

Calculating (7)+(8)(7)+(8) gives us the equation below.

cos⁡x+cos⁡y=2cos⁡x+y2cos⁡x−y2 \cos x + \cos y= 2\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}

Calculating (7)−(8)(7)-(8) gives us the equation below.

cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2 \cos x - \cos y= -2\sin \dfrac{x+y}{2} \sin \dfrac{x-y}{2}

Substituting back to x≡Ax \equiv A, y≡By \equiv B yields the following result.

sin⁡A+sin⁡B= 2sin⁡A+B2cos⁡A−B2sin⁡A−sin⁡B= 2cos⁡A+B2sin⁡A−B2cos⁡A+cos⁡B= 2cos⁡A+B2cos⁡A−B2cos⁡A−cos⁡B= −2sin⁡A+B2sin⁡A−B2 \begin{align*} \sin A + \sin B =&\ 2 \sin \dfrac{A+B}{2} \cos \dfrac{A-B}{2} \\ \sin A - \sin B =&\ 2 \cos \dfrac{A+B}{2} \sin \dfrac{A-B}{2} \\ \cos A + \cos B =&\ 2 \cos \dfrac{A+B}{2} \cos \dfrac{A-B}{2} \\ \cos A - \cos B =&\ -2 \sin \dfrac{A+B}{2} \sin \dfrac{A-B}{2} \end{align*}

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Product-to-Sum Identities

sin⁡Acos⁡B= 12[sin⁡(A+B)+sin⁡(A−B)]cos⁡Asin⁡B= 12[sin⁡(A+B)−sin⁡(A−B)]cos⁡Acos⁡B= 12[cos⁡(A+B)+cos⁡(A−B)]sin⁡Asin⁡B= −12[cos⁡(A+B)−cos⁡(A−B)] \begin{align*} \sin A \cos B =&\ \dfrac{1}{2} \left[ \sin(A+B) + \sin (A-B) \right] \\ \cos A \sin B =&\ \dfrac{1}{2} \left[ \sin(A+B) - \sin (A-B) \right] \\ \cos A \cos B =&\ \dfrac{1}{2} \left[ \cos(A+B) + \cos (A-B) \right] \\ \sin A \sin B =&\ -\dfrac{1}{2} \left[ \cos(A+B) - \cos (A-B) \right] \end{align*}

Derivation

Simple derivation from the trigonometric addition formulas. sin⁡(A+B)= sin⁡Acos⁡B+cos⁡Asin⁡Bsin⁡(A−B)= sin⁡Acos⁡B−cos⁡Asin⁡Bcos⁡(A+B)= cos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡(A−B)= cos⁡Acos⁡B+sin⁡Asin⁡B \begin{align} \sin (A+B) =&\ \sin A \cos B + \cos A \sin B \\ \sin (A-B) =&\ \sin A \cos B - \cos A \sin B \\ \cos (A+B) =&\ \cos A \cos B - \sin A \sin B \\ \cos (A-B) =&\ \cos A \cos B + \sin A \sin B \end{align}

Calculating (9)+(10)(9) + (10) gives the following.

sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B  ⟹  sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)] \sin (A+B) + \sin(A-B) = 2\sin A \cos B \\ \implies \sin A \cos B = \dfrac{1}{2} \left[ \sin(A+B) + \sin (A-B) \right]

Calculating (9)−(10)(9) - (10) gives the following.

sin⁡(A+B)−sin⁡(A−B)=2cos⁡Asin⁡B  ⟹  cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)] \sin (A+B) - \sin(A-B) = 2\cos A \sin B \\ \implies \cos A \sin B = \dfrac{1}{2} \left[ \sin(A+B) - \sin (A-B) \right]

Calculating (11)+(12)(11) + (12) gives the following.

cos⁡(A+B)+cos⁡(A−B)=2cos⁡Acos⁡B  ⟹  cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)] \cos (A+B) + \cos (A-B) = 2 \cos A \cos B \\ \implies \cos A \cos B = \dfrac{1}{2} \left[ \cos(A+B) + \cos (A-B) \right]

Calculating (11)−(12)(11) - (12) gives the following.

cos⁡(A+B)−cos⁡(A−B)=−2sin⁡Asin⁡B  ⟹  sin⁡Asin⁡B=−12[cos⁡(A+B)−cos⁡(A−B)] \cos (A+B) - \cos (A-B) =-2 \sin A \sin B \\ \implies \sin A \sin B = -\dfrac{1}{2} \left[ \cos(A+B) - \cos (A-B) \right]

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