Let S = [0,2) : () + () = 0.

Question

Let $S = \{	heta \in [0,2\pi) : 	an(\pi\cos	heta) + 	an(\pi\sin	heta) = 0\}$. Then $\displaystyle\sum_{	heta \in S}\sin\left(2	heta+\dfrac\pi4ight)$ is equal to \_\_\_\_\_\_\_\_\_\_\_

Answer: 2

Accepted Answer

$$\tan(\pi\cos\theta) = -\tan(\pi\sin\theta) = \tan(-\pi\sin\theta) \implies \pi\cos\theta = n\pi - \pi\sin\theta$$

$$\sin\theta+\cos\theta = n,\quad n \in \mathbb{Z}$$

Since $-\sqrt2 \le \sin\theta+\cos\theta \le \sqrt2$, the possible integer values are $n = -1, 0, 1$. Solving for each case over $[0,2\pi)$ gives:

$$\theta \in \left\{0,\ \frac\pi2,\ \frac{3\pi}4,\ \frac{7\pi}4,\ \frac{3\pi}2,\ \pi\right\}$$

$$\sum_{\theta \in S}\sin\left(2\theta+\frac\pi4\right) = 2(0) + 4\left(\frac12\right) = 2$$

Solutions · 1