Find cos x + cos(x + 2π/3) + cos(x + 4π/3) and sin x + sin(x + 2π/3) + sin(x + 4π/3).
Solution
Using cos(A+B) = cos A cos B - sin A sin B, we have cos(x + 2π/3) = -(1/2) cos x + (√3)/2 sin x, cos(x + 4π/3) = -(1/2) cos x - (√3)/2 sin x. Hence cos x + cos(x + 2π/3) + cos(x + 4π/3) = 0. Similarly, sin(x + 2π/3) = -1/2 sin x + (√3)/2 cos x, sin(x + 4π/3) = -1/2 sin x - (√3)/2 cos x, so sin x + sin(x + 2π/3) + sin(x + 4π/3) = 0.
Thanks to Suat Namli
© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected/updated 7 Mar 04