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Algebra Difficulty 4.9 AIME Prove it Croatia

Prove the identity
cos3x3+cos3x+2π3+cos3x+4π3=34cosx \cos^3 \frac{x}{3} + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x
for every real number xx.

Solution

Using the identity
cost=4cos3t33cost3 \cos t = 4 \cos^3 \frac{t}{3} - 3 \cos \frac{t}{3}
for t=xt = x, t=x+2πt = x + 2\pi and t=x+4πt = x + 4\pi, summing the relations and using the periodicity of cosine, we obtain
3cosx=4(cos3x3+cos3x+2π3+cos3x+4π3)3(cosx3+cosx+2π3+cosx+4π3) 3 \cos x = 4 \left( \cos^3 \frac{x}{3} + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} \right) - 3 \left( \cos \frac{x}{3} + \cos \frac{x+2\pi}{3} + \cos \frac{x+4\pi}{3} \right)
Transforming the expression in the second parenthesis yields
cosx3+cosx+2π3+cosx+4π3=2cos2π3cosx+2π3+cosx+2π3=(2cos2π3+1)cosx+2π3=0. \cos \frac{x}{3} + \cos \frac{x+2\pi}{3} + \cos \frac{x+4\pi}{3} = 2 \cos \frac{2\pi}{3} \cos \frac{x+2\pi}{3} + \cos \frac{x+2\pi}{3} \\ = \left( 2 \cos \frac{2\pi}{3} + 1 \right) \cos \frac{x+2\pi}{3} = 0.
Finally, we conclude
cos3x3+cos3x+2π3+cos3x+4π3=34cosx. \cos^3 \frac{x}{3} + \cos^3 \frac{x+2\pi}{3} + \cos^3 \frac{x+4\pi}{3} = \frac{3}{4} \cos x.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.