A-Level Maclaurin series: worked solution

Question

Triangle ABC has AC = 1, with BAC measuring pi/3 and ABC measuring (pi/6 + theta), both in radians. (a) Show BC = sqrt(3)/(cos theta + sqrt(3) sin theta). [2] (b) For theta small enough to drop theta^3 and higher, show BC approx sqrt(3)(1 + a theta + b theta^2), finding a and b. [3]

(a) In triangle ABC, side BC is opposite angle BAC and side AC is opposite angle ABC, so the Sine Rule pairs them: BC / sin(BAC) = AC / sin(ABC) Rearranging for BC and putting in AC = 1, BAC = pi/3, ABC = pi/6 + theta: BC = AC x sin(BAC) / sin(ABC) = sin(pi/3) / sin(pi/6 + theta) Now expand the denominator with the compound-angle formula sin(P + Q) = sin P cos Q + cos P sin Q: sin(pi/6 + theta) = sin(pi/6) cos(theta) + cos(pi/6) sin(theta) = (1/2) cos(theta) + (sqrt(3)/2) sin(theta) = (1/2)(cos(theta) + sqrt(3) sin(theta)) And sin(pi/3) = sqrt(3)/2, so BC = (sqrt(3)/2) / [ (1/2)(cos(theta) + sqrt(3) sin(theta)) ] The halves cancel: BC = sqrt(3) / (cos(theta) + sqrt(3) sin(theta)) as required. (b) Use the standard small-angle (Maclaurin) series, keeping terms as far as theta² and discarding theta³ and higher: cos(theta) = 1 - theta²/2 + ... sin(theta) = theta - theta³/6 + ... , so sin(theta) is theta to this order So the denominator is cos(theta) + sqrt(3) sin(theta) = 1 + sqrt(3) theta - theta²/2 + ... Write this as 1 + u, where u = sqrt(3) theta - theta²/2 BC is sqrt(3) divided by that, so expand (1 + u)^(-1) using the binomial series (1 + u)^(-1) = 1 - u + u² - u³ + ... Work out how far each power of u reaches: u = sqrt(3) theta - theta²/2 u² = (sqrt(3) theta)² + (a cross term in theta³) + ... = 3 theta² + (theta³ and higher) u³ and beyond are theta³ and higher, so all are dropped. Therefore (1 + u)^(-1) = 1 - (sqrt(3) theta - theta²/2) + 3 theta² + ... = 1 - sqrt(3) theta + theta²/2 + 3 theta² = 1 - sqrt(3) theta + (7/2) theta² Multiplying by sqrt(3): BC = sqrt(3)(1 - sqrt(3) theta + (7/2) theta²) Comparing with the given form sqrt(3)(1 + a theta + b theta²): a = -sqrt(3) and b = 7/2