← Trigonometric Functions (ME-T1, T2, T3)

NSWMaths Extension 1Syllabus dot point

How do we convert between products and sums of trigonometric functions?

Use the product-to-sum and sum-to-product identities to simplify trigonometric expressions and integrals

A focused answer to the HSC Maths Extension 1 dot point on product-to-sum and sum-to-product identities. The four product-to-sum formulas, their sum-to-product converses, derivation from sum and difference, and use in integration, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to convert products like $\sin A \cos B$ into sums (and vice versa) using the standard product-to-sum and sum-to-product identities. These are essential for integrating products of trig functions.

The answer

The four product-to-sum identities

Derived by adding or subtracting the sum and difference identities:

\sin A \cos B = \tfrac{1}{2}\bigl[\sin(A + B) + \sin(A - B)\bigr],

\cos A \sin B = \tfrac{1}{2}\bigl[\sin(A + B) - \sin(A - B)\bigr],

\cos A \cos B = \tfrac{1}{2}\bigl[\cos(A - B) + \cos(A + B)\bigr],

\sin A \sin B = \tfrac{1}{2}\bigl[\cos(A - B) - \cos(A + B)\bigr].

The mnemonic: " $\sin \cos$ produces $\sin$ of sum and difference, $\cos \cos$ produces $\cos$ of sum and difference (plus), $\sin \sin$ produces $\cos$ of difference minus $\cos$ of sum."

Derivation

Add $\sin(A + B) = \sin A \cos B + \cos A \sin B$ and $\sin(A - B) = \sin A \cos B - \cos A \sin B$ . The $\cos A \sin B$ terms cancel: $\sin(A + B) + \sin(A - B) = 2 \sin A \cos B$ . Divide by $2$ to get the first identity.

The other three identities are derived analogously.

Sum-to-product (the converses)

Let $A + B = P$ and $A - B = Q$ , so $A = \frac{P + Q}{2}$ and $B = \frac{P - Q}{2}$ . Substituting:

\sin P + \sin Q = 2 \sin \frac{P + Q}{2} \cos \frac{P - Q}{2},

\sin P - \sin Q = 2 \cos \frac{P + Q}{2} \sin \frac{P - Q}{2},

\cos P + \cos Q = 2 \cos \frac{P + Q}{2} \cos \frac{P - Q}{2},

\cos P - \cos Q = -2 \sin \frac{P + Q}{2} \sin \frac{P - Q}{2}.

Why these identities matter

Three main uses appear in HSC questions.

Integration of products. $\int \sin 3 x \cos 5 x \, dx$ has no antiderivative as written. Convert to $\tfrac{1}{2}[\sin 8 x + \sin(-2 x)] = \tfrac{1}{2}[\sin 8 x - \sin 2 x]$ , which integrates immediately.
Simplification. Sums of sines or cosines can become products, often revealing factorable structure.
Solving equations. Equations like $\sin x + \sin 3 x = 0$ become tractable once converted to a product.

Worked example

Product to sum

Express $\sin 3 x \cos x$ as a sum.

$\sin A \cos B = \tfrac{1}{2}[\sin(A + B) + \sin(A - B)]$ with $A = 3 x$ , $B = x$ .

$= \tfrac{1}{2}[\sin 4 x + \sin 2 x]$ .

Integrate a product

Evaluate $\int 2 \sin 3 x \cos x \, dx$ .

From above, $2 \sin 3 x \cos x = \sin 4 x + \sin 2 x$ .

$\int (\sin 4 x + \sin 2 x) \, dx = -\frac{1}{4} \cos 4 x - \frac{1}{2} \cos 2 x + C$ .

Sum to product

Simplify $\sin 5 x + \sin 3 x$ .

$P = 5 x$ , $Q = 3 x$ . $\frac{P + Q}{2} = 4 x$ , $\frac{P - Q}{2} = x$ .

$\sin 5 x + \sin 3 x = 2 \sin 4 x \cos x$ .

Solve using sum-to-product

Solve $\cos 5 x - \cos x = 0$ for $x \in [0, \pi]$ .

Use $\cos P - \cos Q = -2 \sin \frac{P + Q}{2} \sin \frac{P - Q}{2}$ .

$-2 \sin 3 x \sin 2 x = 0$ , so $\sin 3 x = 0$ or $\sin 2 x = 0$ .

$\sin 3 x = 0$ : $3 x = n \pi$ , $x = \frac{n \pi}{3}$ . In $[0, \pi]$ : $0, \frac{\pi}{3}, \frac{2 \pi}{3}, \pi$ .

$\sin 2 x = 0$ : $2 x = n \pi$ , $x = \frac{n \pi}{2}$ . In $[0, \pi]$ : $0, \frac{\pi}{2}, \pi$ .

Union: $0, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2 \pi}{3}, \pi$ .

Product of two cosines

Express $\cos 4 x \cos 2 x$ as a sum.

$\cos A \cos B = \tfrac{1}{2}[\cos(A - B) + \cos(A + B)]$ .

$= \tfrac{1}{2}[\cos 2 x + \cos 6 x]$ .

Common traps

Sign mistake in $\cos P - \cos Q$: The sum-to-product identity has a minus sign in front of the $2$ : $\cos P - \cos Q = -2 \sin \frac{P + Q}{2} \sin \frac{P - Q}{2}$ .
Mixing up product-to-sum signs: IMATH_3 is the difference of cosines, in that order. Get the order wrong and you flip the sign.
Using $A - B$ when $A < B$: IMATH_6 uses $A - B$ even if that is negative. $\sin(-\theta) = -\sin \theta$ handles it cleanly.
Forgetting the factor of $\tfrac{1}{2}$: Every product-to-sum identity has a $\tfrac{1}{2}$ out the front; every sum-to-product has a $2$ .
Algebra inside the identity: IMATH_12 and $A - B$ must be evaluated in radians or degrees consistently. Mix the two and the answer is garbage.