1977 AHSME Problems/Problem 16

Problem 16

If $i^2 = -1$ , then the sum $\cos{45^\circ} + i\cos{135^\circ} + \cdots + i^n\cos{(45 + 90n)^\circ} + \cdots + i^{40}\cos{3645^\circ}$ equals

$\text{(A)}\ \frac{\sqrt{2}}{2} \qquad \text{(B)}\ -10i\sqrt{2} \qquad \text{(C)}\ \frac{21\sqrt{2}}{2} \qquad\\ \text{(D)}\ \frac{\sqrt{2}}{2}(21 - 20i) \qquad \text{(E)}\ \frac{\sqrt{2}}{2}(21 + 20i)$

Solution

Solution by e_power_pi_times_i

Notice that the sequence repeats every $4$ terms. These $4$ terms are $\cos(45^\circ)$ , $i\cos(135^\circ)$ , $-\cos(225^\circ)$ , and $-i\cos(315^\circ)$ . This is because $\cos(\theta) = \cos(360^\circ - \theta)$ . Also $\cos(\theta) = -\cos(180^\circ - \theta)$ , so the $4$ terms are just $\cos(45^\circ)$ , $-i\cos(45^\circ)$ , $\cos(45^\circ)$ , and $-i\cos(45^\circ)$ , which adds up to $\sqrt{2}-i\sqrt{2}$ . The sequence repeats $10$ times, and an extra $i^{40}\cos(3645^\circ)$ . The sum is $10(\sqrt{2}-i\sqrt{2})+\dfrac{\sqrt{2}}{2} = \dfrac{21\sqrt{2}}{2} - 10i\sqrt{2} = \boxed{\text{(D) }\dfrac{\sqrt{2}}{2}(21-20i)}$ .

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1977 AHSME Problems/Problem 16

Problem 16

Solution