If A + B + C = 180°, Prove That sin 2A + sin 2B − sin 2C = 4 cos A cos B sin C

A standard trigonometric identity problem for a triangle’s angles, not tied to a specific exam paper.

Problem

If $A+B+C=180°$, prove that $\sin2A+\sin2B-\sin2C=4\cos A\cos B\sin C$.

Key insight. Since $A,B,C$ are the angles of a triangle, $A+B=180°-C$ and $C=180°-(A+B)$ — two versions of the same supplementary relationship. Applying the sum-to-product formula to $\sin2A+\sin2B$ first, then again to what’s left after subtracting $\sin2C$, uses each version of that relationship exactly once.

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Approach

Apply the sum-to-product identity $\sin X+\sin Y=2\sin\left(\frac{X+Y}2\right)\cos\left(\frac{X-Y}2\right)$ to $\sin2A+\sin2B$, then use $A+B=180°-C$ to simplify. Subtract $\sin2C=2\sin C\cos C$ and factor out $2\sin C$. What remains is a sum of two cosines, which a second sum-to-product application — this time using $C=180°-(A+B)$ — reduces to $4\cos A\cos B$.

Solution

Step 1 — Apply sum-to-product to sin2A + sin2B

$$\sin2A+\sin2B = 2\sin(A+B)\cos(A-B)$$

Step 2 — Use A + B = 180° − C

$$\sin(A+B) = \sin(180°-C) = \sin C$$

$$\sin2A+\sin2B = 2\sin C\cos(A-B)$$

Step 3 — Subtract sin2C and factor

$$\sin2A+\sin2B-\sin2C = 2\sin C\cos(A-B) – 2\sin C\cos C = 2\sin C\left[\cos(A-B)-\cos C\right]$$

Step 4 — Rewrite cos C using C = 180° − (A+B)

$$\cos C = \cos\left(180°-(A+B)\right) = -\cos(A+B)$$

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

Step 5 — Apply sum-to-product to the cosine sum

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

Step 6 — Combine everything

$$\sin2A+\sin2B-\sin2C = 2\sin C\cdot2\cos A\cos B = 4\cos A\cos B\sin C$$

This matches the right-hand side exactly, completing the proof. $\blacksquare$

Answer

Common mistakes

• Using $A+B=180°-C$ and $C=180°-(A+B)$ inconsistently, or skipping the second substitution. Both versions of the supplementary relationship are needed — one to simplify $\sin(A+B)$ early on, the other to convert $\cos C$ into $-\cos(A+B)$ later. Missing either step leaves the proof stuck with mismatched terms.
• Forgetting the sign flip when writing $\cos(180°-\phi)=-\cos\phi$. This single sign is what converts a subtraction ($\cos(A-B)-\cos C$) into an addition ($\cos(A-B)+\cos(A+B)$), which is essential for the second sum-to-product step to apply cleanly.

Practise next

• If $A+B+C=180°$, prove that $\cos2A+\cos2B+\cos2C=-1-4\cos A\cos B\cos C$, using the same pair of sum-to-product substitutions.