Maths Olympiad Prep

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Problem 1358

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Algebra Difficulty 5.1 Find the answer HMMT February

Simplify the expression: (cos⁡2π3+isin⁡2π3)6+(cos⁡4π3+isin⁡4π3)6\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)^{6} + \left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)^{6} using DeMoivre's Theorem.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

We apply DeMoivre's Theorem to simplify the first expression to (cos⁡6⋅2π3+sin⁡6⋅2π3)=(cos⁡4π+sin⁡4π)=1+0=1\left(\cos 6 \cdot \frac{2 \pi}{3}+\sin 6 \cdot \frac{2 \pi}{3}\right)=(\cos 4 \pi+\sin 4 \pi)=1+0=1. Similarly, we simplify the second expression to (cos⁡6⋅4π3+sin⁡6⋅4π3)=(cos⁡8π+sin⁡8π)=1+0=1\left(\cos 6 \cdot \frac{4 \pi}{3}+\sin 6 \cdot \frac{4 \pi}{3}\right)=(\cos 8 \pi+\sin 8 \pi)=1+0=1. Thus, the total sum is 1+1=21+1=\mathbf{2}.

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