NDA Maths · Trigonometric Identities

Product-to-Sum & Sum-to-Product

The two conversion families — turn a product of sines/cosines into a sum, or a sum into a product — plus the telescoping product chains and the conditional identities for A + B + C = 90° or 180°.

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Why this matters

These conversions are what make otherwise-intractable products like 8 cos 10° cos 20° cos 40° or sums like cos 48° − cos 12° collapse to a clean value. Choosing the right direction (product→sum vs sum→product) is the whole decision.

Test yourself — a quick 12-question Trigonometric Identities recall quiz, instant score.

Concept 1 of 4: Product-to-sum formulas

Turn a product of two sines/cosines into a sum or difference — derived directly by adding and subtracting the compound-angle formulas. Use this when you have a product and want it to telescope or cancel.

Definition

$2\sin A\cos B=\sin(A+B)+\sin(A-B)$ .
$2\cos A\sin B=\sin(A+B)-\sin(A-B)$ .
$2\cos A\cos B=\cos(A+B)+\cos(A-B)$ .
$2\sin A\sin B=\cos(A-B)-\cos(A+B)$ .
The difference-of-squares pair (compound formulas multiplied, not added): $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B=\cos^2B-\cos^2A$ and $\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B$ . So $\sin^2 75°-\sin^2 15°=\sin 90°\sin 60°=\tfrac{\sqrt3}{2}$ in one step.

The four product-to-sum identities

2\sin A\cos B=\sin(A+B)+\sin(A-B),\qquad 2\cos A\cos B=\cos(A+B)+\cos(A-B),\qquad 2\cos A\sin B=\sin(A+B)-\sin(A-B),\qquad 2\sin A\sin B=\cos(A-B)-\cos(A+B)

Worked example

Evaluate

2\sin 75°\cos 15°

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q40Moderate

Example 1 · Trigonometric Identities · Product-to-Sum and Sum-to-Product Identities

What is

\dfrac{1}{\sin 10°} - \dfrac{\sqrt{3}}{\cos 10°}

equal to?

$2\sin A\sin B=\cos(A-B)-\cos(A+B)$ — cosines, and the difference comes FIRST

The two-sine product is the most error-prone: it converts to cosines, not sines, and the order is

\cos(A-B)-\cos(A+B)

(difference minus sum). Students write

\cos(A+B)-\cos(A-B)

and get the whole sign backwards. Contrast

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

, which is a plus. Tip: the

2\sin A\sin B

one is the only product-to-sum identity with a leading minus.

Drill 7 more on product-to-sum formulas

Concept 2 of 4: Sum-to-product formulas

The reverse direction: a sum or difference of two sines/cosines becomes a product. Use this when you want a common factor to cancel or a ratio to simplify to a single tangent.

Definition

$\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}$ ; $\;\sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2}$ .
$\cos C+\cos D=2\cos\tfrac{C+D}{2}\cos\tfrac{C-D}{2}$ ; $\;\cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}$ .

Corollary: $\dfrac{\sin C+\sin D}{\cos C+\cos D}=\tan\tfrac{C+D}{2}$ .

The four sum-to-product identities

\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2},\qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \cos C+\cos D=2\cos\tfrac{C+D}{2}\cos\tfrac{C-D}{2}

Worked example

Simplify

\dfrac{\sin 5x-\sin 3x}{\cos 5x+\cos 3x}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q28Moderate

Example 2 · Trigonometric Identities · Product-to-Sum and Sum-to-Product Identities

What is the value of

\cos 48°-\cos 12°

Half-SUM and half-DIFFERENCE go in fixed slots — don't swap them

\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2}

, the half-sum $\tfrac{C+D}{2}$ sits inside the leading function and the half-difference

\tfrac{C-D}{2}

inside the trailing one. Students swap them, or use

C-D

and

C+D

without halving. Also remember

\cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2}

carries a leading minus (so if

C>D

and both are acute, the difference is negative).

Drill 10 more on sum-to-product formulas

Concept 3 of 4: Telescoping products of cosines/sines

A chain like cos 10° cos 20° cos 40° collapses by repeatedly using

2\sin\theta\cos\theta=\sin 2\theta

: introduce a sine, and each cosine doubles the angle until the product telescopes.

Definition

Multiply and divide by $2\sin(\text{smallest angle})$ , then apply $2\sin\theta\cos\theta=\sin 2\theta$ repeatedly. General result: $\cos\theta\cos 2\theta\cos 4\theta\cdots\cos 2^{n-1}\theta=\dfrac{\sin 2^n\theta}{2^n\sin\theta}$ . Triple products like $\sin\theta\sin(60°-\theta)\sin(60°+\theta)=\tfrac14\sin 3\theta$ also appear. Sums in arithmetic angles telescope too, by product-to-sum with the common difference: $\sum_{k=0}^{n-1}\sin(a+kd)=\dfrac{\sin\frac{nd}{2}}{\sin\frac d2}\,\sin\!\Big(a+\frac{(n-1)d}{2}\Big)$ , and the same prefactor with $\cos$ for a cosine series. For a symmetric run like $\sin 10°+\sin 20°+\dots+\sin 170°$ pairing $\sin\theta=\sin(180°-\theta)$ is faster.

Worked example

Evaluate

\cos 20°\cos 40°\cos 80°

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2020 · Q27Hard

Example 3 · Trigonometric Identities · Product-to-Sum and Sum-to-Product Identities

What is the value of

8\cos 10°\cdot\cos 20°\cdot\cos 40°

Drill 5 more on telescoping products of cosines/sines

Concept 4 of 4: Conditional identities (A + B + C = 90° or 180°)

When three angles sum to 90° or 180°, special identities kick in — the staple results for triangle-angle problems. Recognising the angle-sum condition is the trigger.

Definition

$A+B+C=180°$ : $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ ; $\;\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ .
$A+B+C=90°$ : $\tan A\tan B+\tan B\tan C+\tan C\tan A=1$ ; $\;\cot A+\cot B+\cot C=\cot A\cot B\cot C$ .

The two signature conditional identities

A+B+C=\pi:\ \tan A+\tan B+\tan C=\tan A\tan B\tan C

Worked example

A+B+C=180°

and

\tan A=1,\tan B=2

, find

\tan C

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q70Moderate

Example 4 · Trigonometric Identities · Product-to-Sum and Sum-to-Product Identities

What is

\tan25°\tan15° + \tan15°\tan50° + \tan25°\tan50°

equal to ?

$\tan A+\tan B+\tan C=\tan A\tan B\tan C$ only when $A+B+C=180°$

These are conditional identities — they hold only under the stated angle-sum, not for arbitrary angles. The

\sum\tan=\prod\tan

relation needs

A+B+C=180°

; the

\sum\tan A\tan B=1

relation needs

A+B+C=90°

. Applying the wrong one (or applying either to angles that don't sum correctly) is the trap. Verify the angle-sum condition before invoking the identity.

Drill 1 more on conditional identities (a + b + c = 90° or 180°)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Product-to-sum formulas
The four product-to-sum identities
$2\sin A\cos B=\sin(A+B)+\sin(A-B),\qquad 2\cos A\cos B=\cos(A+B)+\cos(A-B),\qquad 2\cos A\sin B=\sin(A+B)-\sin(A-B),\qquad 2\sin A\sin B=\cos(A-B)-\cos(A+B)$
Sum-to-product formulas
The four sum-to-product identities
$\sin C+\sin D=2\sin\tfrac{C+D}{2}\cos\tfrac{C-D}{2},\qquad \cos C-\cos D=-2\sin\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \sin C-\sin D=2\cos\tfrac{C+D}{2}\sin\tfrac{C-D}{2},\qquad \cos C+\cos D=2\cos\tfrac{C+D}{2}\cos\tfrac{C-D}{2}$
Conditional identities (A + B + C = 90° or 180°)
The two signature conditional identities
$A+B+C=\pi:\ \tan A+\tan B+\tan C=\tan A\tan B\tan C$

Watch out for (3)

$2\sin A\sin B=\cos(A-B)-\cos(A+B)$ — cosines, and the difference comes FIRST
→ Product-to-sum formulas
Half-SUM and half-DIFFERENCE go in fixed slots — don't swap them
→ Sum-to-product formulas
$\tan A+\tan B+\tan C=\tan A\tan B\tan C$ only when $A+B+C=180°$
→ Conditional identities (A + B + C = 90° or 180°)

Test yourself on Trigonometric Identities

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