Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 03 / 11Trigonometry

Transformation Formulas: Sum-to-Product, Product-to-Sum and Sums of Trigonometric Series

Converting sums of sines and cosines into products and products into sums, their use in simplification and in solving equations, the sum of sin and cos of angles in arithmetic progression, and telescoping trigonometric sums.

Adding and subtracting the compound angle formulas gives the product-to-sum formulas: 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A + B) + \sin(A - B), 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B = \sin(A + B) - \sin(A - B), 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B = \cos(A + B) + \cos(A - B), 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B). Setting C=A+BC = A + B and D=A−BD = A - B reverses them into the sum-to-product formulas: sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2\sin C + \sin D = 2\sin\frac{C + D}{2}\cos\frac{C - D}{2}, sin⁡C−sin⁡D=2cos⁡C+D2sin⁡C−D2\sin C - \sin D = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2}, cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C + \cos D = 2\cos\frac{C + D}{2}\cos\frac{C - D}{2}, cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2}. The minus sign in the last formula is the most common source of errors. When to use which. Sum-to-product turns sums into products, which is what you want when solving an equation (a product is zero iff a factor is zero) or when simplifying a ratio of sums, where common factors cancel. Product-to-sum turns products into sums, which is what you want before integrating or summing a series. Sum of sines and cosines of angles in AP. For nn angles α,α+β,…,α+(n−1)β\alpha, \alpha + \beta, \ldots, \alpha + (n - 1)\beta: ∑k=0n−1sin⁡(α+kβ)=sin⁡nβ2sin⁡β2sin⁡(α+(n−1)β2)\sum_{k=0}^{n-1}\sin(\alpha + k\beta) = \frac{\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}\sin\left(\alpha + \frac{(n - 1)\beta}{2}\right), ∑k=0n−1cos⁡(α+kβ)=sin⁡nβ2sin⁡β2cos⁡(α+(n−1)β2)\sum_{k=0}^{n-1}\cos(\alpha + k\beta) = \frac{\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}\cos\left(\alpha + \frac{(n - 1)\beta}{2}\right). Memory aid: sin⁡(n×half the difference)sin⁡(half the difference)\frac{\sin(n \times \text{half the difference})}{\sin(\text{half the difference})} times the sine (or cosine) of the average of the first and last angles. Proof: multiply the sum by 2sin⁡β22\sin\frac{\beta}{2}; each term becomes a difference of cosines (or sines) by product-to-sum, and the sum telescopes. In complex form, it is the real or imaginary part of a geometric series in eiβe^{i\beta}. A consequence: if nβn\beta is a multiple of 2π2\pi (and β\beta is not), the sum is 00, for example cos⁡2π7+cos⁡4π7+⋯+cos⁡14π7=0\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cdots + \cos\frac{14\pi}{7} = 0, reflecting that the nnth roots of unity sum to zero. Telescoping. Products like sin⁡xsin⁡kx\sin x\sin kx or ratios like sin⁡xcos⁡kxcos⁡(k+1)x=tan⁡(k+1)x−tan⁡kx\frac{\sin x}{\cos kx\cos(k + 1)x} = \tan(k + 1)x - \tan kx let sums collapse. The general idea: convert each term into a difference f(k+1)−f(k)f(k + 1) - f(k). Key Takeaways: - 2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A + B) + \sin(A - B); 2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B). - cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C - \cos D = -2\sin\frac{C + D}{2}\sin\frac{C - D}{2} (note the minus). - Sum-to-product for solving equations and simplifying ratios; product-to-sum for integrating and summing. - ∑sin⁡(α+kβ)=sin⁡(nβ/2)sin⁡(β/2)sin⁡(mean angle)\sum\sin(\alpha + k\beta) = \frac{\sin(n\beta/2)}{\sin(\beta/2)}\sin(\text{mean angle}). - If nβ=2πmn\beta = 2\pi m, the AP sums vanish. - Pitfall: applying the AP-sum formula when the angles are not in arithmetic progression. Brain-Teaser 1: Simplify sin⁡7x+sin⁡5xcos⁡7x+cos⁡5x\frac{\sin 7x + \sin 5x}{\cos 7x + \cos 5x}. Solution: Numerator =2sin⁡6xcos⁡x= 2\sin 6x\cos x; denominator =2cos⁡6xcos⁡x= 2\cos 6x\cos x. The ratio is tan⁡6x\tan 6x (where defined). The common factor cos⁡x\cos x is what sum-to-product exposes. Brain-Teaser 2: Evaluate cos⁡π11+cos⁡3π11+cos⁡5π11+cos⁡7π11+cos⁡9π11\cos\frac{\pi}{11} + \cos\frac{3\pi}{11} + \cos\frac{5\pi}{11} + \cos\frac{7\pi}{11} + \cos\frac{9\pi}{11}. Solution: Five terms in AP with α=π11\alpha = \frac{\pi}{11}, β=2π11\beta = \frac{2\pi}{11}. The sum is sin⁡5π11sin⁡π11cos⁡5π11=sin⁡10π112sin⁡π11=12\frac{\sin\frac{5\pi}{11}}{\sin\frac{\pi}{11}}\cos\frac{5\pi}{11} = \frac{\sin\frac{10\pi}{11}}{2\sin\frac{\pi}{11}} = \frac{1}{2}, since sin⁡10π11=sin⁡π11\sin\frac{10\pi}{11} = \sin\frac{\pi}{11}. A neat half, found without any numerical work. Brain-Teaser 3: Solve sin⁡x+sin⁡3x+sin⁡5x=0\sin x + \sin 3x + \sin 5x = 0. Solution: Group the outer terms: sin⁡5x+sin⁡x=2sin⁡3xcos⁡2x\sin 5x + \sin x = 2\sin 3x\cos 2x. So the equation is sin⁡3x(2cos⁡2x+1)=0\sin 3x(2\cos 2x + 1) = 0. Either sin⁡3x=0\sin 3x = 0, giving x=nπ3x = \frac{n\pi}{3}, or cos⁡2x=−12\cos 2x = -\frac{1}{2}, giving 2x=2nπ±2π32x = 2n\pi \pm \frac{2\pi}{3}, i.e. x=nπ±π3x = n\pi \pm \frac{\pi}{3}. The second family is already contained in the first (nπ±π3n\pi \pm \frac{\pi}{3} are multiples of π3\frac{\pi}{3}), so the complete solution is x=nπ3x = \frac{n\pi}{3}, n∈Zn \in \mathbb{Z}. Grouping terms symmetrically about the middle angle makes the common factor appear.

Practice this concept

Free · no account required · instant solutions

Transformation Formulas: Sum-to-Product, Product-to-Sum and Sums of Trigonometric Series — DPP

4 questionsSee the solution the moment you answer

Practice