O-Level Trigonometric functions, identities and equations: worked solution

Question

Given cos A = 1/2 with 0° < A < 90°, and sin B = −1/√2 with 180° < B < 270°, evaluate cos(A − B) exactly and without a calculator, expressing it as p√2 + q√6 for real p and q.

cos A = 1/2 with A in first quadrant, so A = 60 deg: cos A = 1/2, sin A = sqrt(3)/2. sin B = -1/sqrt(2) with 180 < B < 270 (third quadrant), so cos B is negative: cos B = -sqrt(1 - 1/2) = -1/sqrt(2). cos(A - B) = cos A cos B + sin A sin B = (1/2)(-1/sqrt(2)) + (sqrt(3)/2)(-1/sqrt(2)) = -1/(2 sqrt(2)) - sqrt(3)/(2 sqrt(2)). Rationalise: -1/(2 sqrt(2)) = -sqrt(2)/4; -sqrt(3)/(2 sqrt(2)) = -sqrt(6)/4. So cos(A - B) = -sqrt(2)/4 - sqrt(6)/4 = -(1/4)sqrt(2) - (1/4)sqrt(6). Thus p = -1/4, q = -1/4.