Trigonometric Identities: Edge Cases and Subtle Traps (3)

Question

Prove that $\dfrac{\sin\theta + \sin 3\theta}{\cos\theta + \cos 3\theta} = \tan 2\theta$.

Solution — Step by Step

Why This Works

Sum-to-product identities convert a sum of sines (or cosines) into a product. The product form often has a common factor that cancels — that’s the key. Both numerator and denominator share the factor $2\cos\theta$ here, leaving $\sin 2\theta / \cos 2\theta = \tan 2\theta$.

This pattern generalises: $(\sin A + \sin B)/(\cos A + \cos B) = \tan((A+B)/2)$.

Alternative Method

Expand $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. Substituting and simplifying eventually gives the same result, but with much more algebra. Sum-to-product is faster.

Common Mistake

Sign error in the cosine sum-to-product formula. The minus version is $\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$ — note the negative sign. Sum has cosines; difference has sines, with a sign flip. Mixing these up flips the sign of the whole expression.