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9.4 Sum-to-Product and Product-to-Sum Formulas

Updated March 2026Fiveable Content Team
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📈College Algebra Unit 9 Review

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9.4 Sum-to-Product and Product-to-Sum Formulas

Sum-to-Product and Product-to-Sum Formulas

Sum-to-product and product-to-sum formulas let you rewrite trig expressions in equivalent forms that are easier to work with. A sum like sin⁡A+sin⁡B\sin A + \sin B can become a product, and a product like cos⁡Acos⁡B\cos A \cos B can become a sum. This flexibility is what makes these formulas so useful for simplifying expressions and solving trig equations.

Sum-to-Product Formulas

These formulas convert a sum or difference of two trig functions into a product. Each formula uses the average of the two angles, A+B2\frac{A+B}{2}, and the half-difference, A−B2\frac{A-B}{2}.

  • sin⁡A+sin⁡B=2sin⁡ ⁣(A+B2)cos⁡ ⁣(A−B2)\sin A + \sin B = 2 \sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)
  • sin⁡A−sin⁡B=2cos⁡ ⁣(A+B2)sin⁡ ⁣(A−B2)\sin A - \sin B = 2 \cos\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)
  • cos⁡A+cos⁡B=2cos⁡ ⁣(A+B2)cos⁡ ⁣(A−B2)\cos A + \cos B = 2 \cos\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)
  • cos⁡A−cos⁡B=−2sin⁡ ⁣(A+B2)sin⁡ ⁣(A−B2)\cos A - \cos B = -2 \sin\!\left(\frac{A+B}{2}\right)\sin\!\left(\frac{A-B}{2}\right)

Notice the pattern: sine formulas pair a sine with a cosine, while cosine formulas pair two of the same function. The cosine difference formula has a negative sign out front, which is easy to forget.

Example: Rewrite sin⁡5x+sin⁡3x\sin 5x + \sin 3x as a product.

  1. Identify A=5xA = 5x and B=3xB = 3x.
  2. Compute A+B2=5x+3x2=4x\frac{A+B}{2} = \frac{5x+3x}{2} = 4x and A−B2=5x−3x2=x\frac{A-B}{2} = \frac{5x-3x}{2} = x.
  3. Apply the sine sum formula: sin⁡5x+sin⁡3x=2sin⁡(4x)cos⁡(x)\sin 5x + \sin 3x = 2\sin(4x)\cos(x).

This product form is often much easier to factor or set equal to zero when solving equations.

Product-to-Sum Formulas

These formulas go the other direction: they convert a product of two trig functions into a sum or difference.

  • sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]
  • cos⁡Asin⁡B=12[sin⁡(A+B)−sin⁡(A−B)]\cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]
  • cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)]\cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)]
  • sin⁡Asin⁡B=−12[cos⁡(A+B)−cos⁡(A−B)]\sin A \sin B = -\frac{1}{2}[\cos(A+B) - \cos(A-B)]

A quick way to remember: when both factors are cosines, the result involves cosines. When both are sines, the result also involves cosines but with a negative sign out front. Mixed pairs (sine × cosine) produce sines.

Example: Rewrite cos⁡3θcos⁡5θ\cos 3\theta \cos 5\theta as a sum.

  1. Identify A=3θA = 3\theta and B=5θB = 5\theta.
  2. Apply the cosine-cosine formula: cos⁡3θcos⁡5θ=12[cos⁡(3θ+5θ)+cos⁡(3θ−5θ)]\cos 3\theta \cos 5\theta = \frac{1}{2}[\cos(3\theta+5\theta) + \cos(3\theta-5\theta)].
  3. Simplify: =12[cos⁡8θ+cos⁡(−2θ)]= \frac{1}{2}[\cos 8\theta + \cos(-2\theta)].
  4. Since cos⁡(−x)=cos⁡x\cos(-x) = \cos x, the final answer is 12[cos⁡8θ+cos⁡2θ]\frac{1}{2}[\cos 8\theta + \cos 2\theta].
Sum-to-product formula application, Sum and Difference Identities – Algebra and Trigonometry OpenStax

Solving Equations with These Formulas

The real payoff comes when you use these formulas to solve trig equations. The general strategy is to convert the equation into a form where you can factor or isolate a single trig function.

Steps for solving:

  1. Identify whether the equation contains a sum/difference or a product of trig functions.
  2. Apply the appropriate formula to rewrite the expression.
  3. Set each factor equal to zero (if you now have a product) or simplify further using other identities.
  4. Solve each resulting equation for the variable.
  5. Check your solutions in the original equation, since algebraic manipulation can sometimes introduce extraneous solutions.

Example: Solve sin⁡3x+sin⁡x=0\sin 3x + \sin x = 0.

  1. Apply sum-to-product: 2sin⁡ ⁣(3x+x2)cos⁡ ⁣(3x−x2)=02\sin\!\left(\frac{3x+x}{2}\right)\cos\!\left(\frac{3x-x}{2}\right) = 0, which gives 2sin⁡(2x)cos⁡(x)=02\sin(2x)\cos(x) = 0.
  2. Set each factor to zero: sin⁡(2x)=0\sin(2x) = 0 or cos⁡(x)=0\cos(x) = 0.
  3. sin⁡(2x)=0\sin(2x) = 0 when 2x=nπ2x = n\pi, so x=nπ2x = \frac{n\pi}{2}. And cos⁡(x)=0\cos(x) = 0 when x=π2+nπx = \frac{\pi}{2} + n\pi. The cosine solutions are already included in x=nπ2x = \frac{n\pi}{2}, so the full solution set is x=nπ2x = \frac{n\pi}{2}, where nn is any integer.

Sum-to-product and product-to-sum formulas are actually derived from the angle addition and subtraction identities. You'll sometimes need to combine them with other identities to fully simplify an expression:

  • Double-angle formulas express functions like sin⁡2A\sin 2A or cos⁡2A\cos 2A in terms of sin⁡A\sin A and cos⁡A\cos A.
  • Half-angle formulas express functions of A2\frac{A}{2} in terms of AA.

These come up together often. For instance, after applying a sum-to-product formula you might end up with sin⁡(2x)\sin(2x), which you can expand using the double-angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x if that helps you factor further.