Question

Let $S=\{\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\}$ Then $\displaystyle\sum_{\theta \in s } \sin ^2\left(\theta+\frac{\pi}{4}\right)$ is equal to _______

Answer 1

Correct Answer: 2

To solve the given problem, we need to analyze the set \( S = \{\theta \in [0, 2\pi) : \tan(\pi \cos \theta) + \tan(\pi \sin \theta) = 0\} \). The condition \( \tan(\pi \cos \theta) + \tan(\pi \sin \theta) = 0 \) implies that \( \tan(\pi \cos \theta) = -\tan(\pi \sin \theta) \). This is satisfied when \( \pi \cos \theta \equiv \pi \sin \theta + \frac{\pi}{2} \pmod{\pi} \) or vice versa.

This leads us to analyze: \( \cos \theta = \sin \theta \) or \( \cos \theta = -\sin \theta \). Solving these equations:

• For \( \cos \theta = \sin \theta \), we have \( \theta = \frac{\pi}{4}, \frac{5\pi}{4} \).
• For \( \cos \theta = -\sin \theta \), we have \( \theta = \frac{3\pi}{4}, \frac{7\pi}{4} \).

Thus, \( S = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} \).

Next, compute \(\sum_{\theta \in S} \sin^2\left(\theta+\frac{\pi}{4}\right)\):

• For \( \theta = \frac{\pi}{4} \), \(\sin^2\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = \sin^2\left(\frac{\pi}{2}\right) = 1\).
• For \( \theta = \frac{3\pi}{4} \), \(\sin^2\left(\frac{3\pi}{4}+\frac{\pi}{4}\right) = \sin^2(\pi) = 0\).
• For \( \theta = \frac{5\pi}{4} \), \(\sin^2\left(\frac{5\pi}{4}+\frac{\pi}{4}\right) = \sin^2\left(\frac{3\pi}{2}\right) = 1\).
• For \( \theta = \frac{7\pi}{4} \), \(\sin^2\left(\frac{7\pi}{4}+\frac{\pi}{4}\right) = \sin^2(2\pi) = 0\).

Summing these results gives \( 1 + 0 + 1 + 0 = 2 \).

Therefore, the value of \(\sum_{\theta \in S} \sin^2\left(\theta+\frac{\pi}{4}\right)\) is 2, which lies within the specified range (2, 2).