Sum to Product Formulas — Trig Identities, Proof

Sum to Product Formulas — Trig Identities, Proof

The sum to product formula family converts a sum or difference of two sines (or two cosines) into a product of one sine and one cosine — four identities that turn sin⁡75°+sin⁡15° into a single product expression solvable in one step. This article gives the four formulas, the proof via sum-and-difference identities, three worked examples in degrees and radians, and the common mistake of mixing up the half-sum and half-difference angles.

Four Formulas That Turn Stubborn Sums Into Easy Products

The sum to product formula family is a set of four trigonometric identities:

Each rewrites a sum or difference of two sinusoids as twice the product of a sine (or cosine) of the half-sum and a sine (or cosine) of the half-difference.

The Four Formulas

[ \sin A + \sin B = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right) ]

[ \sin A - \sin B = 2 \cos\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right) ]

[ \cos A + \cos B = 2 \cos\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right) ]

[ \cos A - \cos B = -2 \sin\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right) ]

The pattern: the half-sum angle goes inside the function on the left of the product; the half-difference angle goes inside the function on the right.

Quick facts

Double-Anchoring — Right Triangle and Unit Circle

The proof comes directly from the sum-and-difference identities, which themselves rest on the unit circle.

Let X=(A+B)/2 and Y=(A−B)/2. Then X+Y=A and X−Y=B. Apply the sine sum and difference formulas:

[ \sin A + \sin B = 2 \sin X \cos Y ]

[ \sin A - \sin B = \sin X \cos Y - \cos X \sin Y ]

Add the two equations:

[ \sin A + \sin B = 2 \sin X \cos Y = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A-B}{2}\right) ]

On the unit circle. When A and B are close, (A−B)/2 is small, thus (\cos((A−B)/2)\approx 1), yielding the average of the two angles' sines, doubled.

Three Worked Examples — Quick, Standard, Stretch

Quick. Evaluate (\sin 75°+\sin 15°) exactly.

Apply (\sin A + \sin B = 2 \sin\left(\dfrac{A+B}{2}\right) \cos\left(\dfrac{A−B}{2}\right)) with A=75° and B=15°:

[ \sin 75° + \sin 15° = 2 \sin\left(\dfrac{75° + 15°}{2}\right) \cos\left(\dfrac{75° - 15°}{2}\right) = 2 \sin 45° \cos 30°. ]

Substitute special-angle values: (\sin 45° = \sqrt{2}/2) and (\cos 30° = \sqrt{3}/2).

[ \sin 75° + \sin 15° = 2 \cdot \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{6}}{2}. ]

Final answer: (\sin 75° + \sin 15° = \dfrac{\sqrt{6}}{2}.)

Standard. Evaluate (\cos 75°−\cos 15°) exactly.

Apply the correct identity:

[ \cos A - \cos B = -2 \sin\left(\dfrac{A+B}{2}\right) \sin\left(\dfrac{A-B}{2}\right). ]

[ \cos 75° - \cos 15° = -2 \sin 45° \sin 30° = -2 \cdot \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = -\dfrac{\sqrt{2}}{2}. ]

Final answer: (\cos 75° - \cos 15° = -\dfrac{\sqrt{2}}{2}.)

Stretch. Express (\sin x + \sin 3x + \sin 5x + \sin 7x) as a product.

Using the sum-to-product identities leads to:

[ 4 \sin 4x \cos 2x \cos x. ]

In radians, the variable is x throughout; in degrees, x could equally be written as x°.

Final answer: (\sin x + \sin 3x + \sin 5x + \sin 7x = 4 \sin 4x \cos 2x \cos x.)

Why the Sum to Product Identities Matter Outside the Classroom

The sum-to-product identities are the algebraic engine behind almost every wave-interference calculation in science and engineering.

Frequently Asked Questions