Product-to-sum and sum-to-product formulas

These formulas let us swap products of trigonometric functions for sums, and vice versa. They are workhorses for evaluating sums like $\sin 1^\circ + \sin 2^\circ + \dots + \sin 89^\circ$ and for integration in Class XII.

Product-to-sum

From the sum and difference formulas, by addition and subtraction:

$2 \sin A \cos B = \sin(A + B) + \sin(A - B)$

$2 \cos A \sin B = \sin(A + B) - \sin(A - B)$

$2 \cos A \cos B = \cos(A - B) + \cos(A + B)$

$2 \sin A \sin B = \cos(A - B) - \cos(A + B)$

These hold for all angles $A, B$.

Sum-to-product

Substituting $A = \dfrac{C + D}{2}$ and $B = \dfrac{C - D}{2}$ in the previous identities gives:

$\sin C + \sin D = 2 \sin\!\frac{C + D}{2} \cos\!\frac{C - D}{2}$

$\sin C - \sin D = 2 \cos\!\frac{C + D}{2} \sin\!\frac{C - D}{2}$

$\cos C + \cos D = 2 \cos\!\frac{C + D}{2} \cos\!\frac{C - D}{2}$

$\cos C - \cos D = -2 \sin\!\frac{C + D}{2} \sin\!\frac{C - D}{2}$

Each rewrites a sum as a product , usually the key step in solving trigonometric equations involving multiple sines/cosines.

Derivation outline

From $\sin(A + B) - \sin(A - B) = 2\cos A \sin B$ (which expands to $\sin A \cos B + \cos A \sin B - \sin A \cos B + \cos A \sin B = 2 \cos A \sin B$). Now let $C = A + B$ and $D = A - B$, so $A = \dfrac{C+D}{2}$ and $B = \dfrac{C - D}{2}$. The identity becomes $\sin C - \sin D = 2 \cos\!\frac{C+D}{2} \sin\!\frac{C - D}{2}.$

The other three are similar.

Worked examples

Example 1. Express $\sin 50^\circ - \sin 10^\circ$ as a product.

$\sin 50 - \sin 10 = 2 \cos 30 \sin 20 = 2 \cdot \dfrac{\sqrt{3}}{2} \cdot \sin 20 = \sqrt{3}\sin 20^\circ$.

Example 2. Show $\cos 75^\circ - \cos 15^\circ = -\dfrac{\sqrt{6} - \sqrt{2}}{2}$... wait, simpler: $\cos 75 - \cos 15 = -2 \sin 45 \sin 30 = -2 \cdot \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = -\dfrac{\sqrt{2}}{2}$.

Example 3. Prove $\cos A + \cos(120^\circ - A) + \cos(120^\circ + A) = 0$.

Group the last two terms by sum-to-product: $\cos(120 - A) + \cos(120 + A) = 2\cos 120 \cos A = 2 \cdot \left(-\tfrac{1}{2}\right) \cos A = -\cos A.$ Adding $\cos A$ gives $0$. $\qed$

Example 4. Show $\sin\theta + \sin 3\theta + \sin 5\theta + \sin 7\theta = 4 \cos\theta \cos 2\theta \sin 4\theta$.

Pair: $(\sin\theta + \sin 7\theta) + (\sin 3\theta + \sin 5\theta) = 2\sin 4\theta \cos 3\theta + 2 \sin 4\theta \cos\theta = 2\sin 4\theta (\cos 3\theta + \cos\theta) = 2\sin 4\theta \cdot 2 \cos 2\theta \cos\theta = 4\sin 4\theta \cos 2\theta \cos\theta$. $\qed$

Example 5 (harder). Evaluate $\cos\dfrac{\pi}{7} \cos\dfrac{2\pi}{7} \cos\dfrac{3\pi}{7}$.

Multiply numerator and denominator by $2\sin\dfrac{\pi}{7}$ and use $2\sin\theta\cos\theta = \sin 2\theta$ repeatedly: $2\sin\tfrac{\pi}{7} \cos\tfrac{\pi}{7} = \sin\tfrac{2\pi}{7}.$ So $\sin\tfrac{2\pi}{7} \cdot \cos\tfrac{2\pi}{7} = \tfrac{1}{2} \sin\tfrac{4\pi}{7}$, and $\tfrac{1}{2}\sin\tfrac{4\pi}{7} \cdot \cos\tfrac{3\pi}{7}$...

Alternative: note $\cos\tfrac{3\pi}{7} = -\cos\tfrac{4\pi}{7}$ (since $\tfrac{3\pi}{7} + \tfrac{4\pi}{7} = \pi$). So our product becomes $\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}(-\cos\tfrac{4\pi}{7})$. Multiply by $2^3 \sin\tfrac{\pi}{7}$ and apply double-angle repeatedly:

$8\sin\tfrac{\pi}{7} \cdot \cos\tfrac{\pi}{7} \cos\tfrac{2\pi}{7} \cos\tfrac{4\pi}{7} = 4 \sin\tfrac{2\pi}{7} \cos\tfrac{2\pi}{7} \cos\tfrac{4\pi}{7} = 2 \sin\tfrac{4\pi}{7} \cos\tfrac{4\pi}{7} = \sin\tfrac{8\pi}{7}.$

Now $\sin\tfrac{8\pi}{7} = \sin(\pi + \tfrac{\pi}{7}) = -\sin\tfrac{\pi}{7}$. So $8 \sin\tfrac{\pi}{7} \cdot \cos\tfrac{\pi}{7} \cos\tfrac{2\pi}{7}\cos\tfrac{4\pi}{7} = -\sin\tfrac{\pi}{7}$, giving $\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{4\pi}{7} = -\tfrac{1}{8}$.

Including the minus sign we removed: $\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{3\pi}{7} = -(-\tfrac{1}{8}) = \tfrac{1}{8}$.

Try it yourself

1. Express $2\sin 5x \cos 3x$ as a sum.
2. Express $\cos 7\theta - \cos 5\theta$ as a product.
3. Prove $\sin A + \sin 3A + \sin 5A = \sin 3A (1 + 2\cos 2A)$.
4. Show $\cos 20^\circ + \cos 100^\circ + \cos 140^\circ = 0$.
5. Find the exact value of $\sin 75^\circ + \sin 15^\circ$.
6. Prove $\sin x + \sin 2x + \sin 3x = \sin 2x (1 + 2\cos x)$.
7. Evaluate $\cos 36^\circ - \cos 72^\circ$. (Hint: classical surd identity.)
8. Express $\sin 7x \cos 4x$ as a sum.
9. Prove $\dfrac{\sin A + \sin 3A}{\cos A + \cos 3A} = \tan 2A$.
10. Show $4 \sin\!\dfrac{\pi}{5} \sin\!\dfrac{2\pi}{5} = \sqrt{5}$. (Use product-to-sum and known cosine values.)
11. Prove $\cos A \cos(60 - A) \cos(60 + A) = \dfrac{1}{4} \cos 3A$.
12. Show $\sin\theta \sin(60 - \theta) \sin(60 + \theta) = \dfrac{1}{4}\sin 3\theta$.

Pitfalls / Tricks

• The product-to-sum formulas have no coefficient $1/2$ on the LHS in this form , you must divide if you isolate the product on one side.
• Watch the sign: $\cos C - \cos D = -2 \sin\dfrac{C+D}{2} \sin\dfrac{C-D}{2}$.
• The "average / half-difference" pattern is universal: every sum-to-product converts $C, D$ into $\tfrac{C+D}{2}, \tfrac{C-D}{2}$.
• Insight. Whenever a problem has a sum like $\sin A + \sin B$ or $\cos C - \cos D$, immediately rewrite it as a product. Half of trigonometric simplification is recognising this pattern.