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

The correct answer is 2
$\tan (\pi \cos \theta )+\tan (\pi \sin \theta )=0$
$\tan (\pi \cos \theta )=-\tan (\pi \sin \theta )$
$\tan (\pi \cos \theta )=\tan (-\pi \sin \theta )$
$\pi \cos \theta =n\pi -\pi \sin \theta$
$\sin \theta +\cos \theta =n\text{where}n\in I$
possible values are $n=0,1$ and $-1$ because
$-\sqrt{2}\le \sin \theta +\cos \theta \le \sqrt{2}$
Now it gives $\theta \in \left\{0,\frac{\pi }{2},\frac{3\pi }{4},\frac{7\pi }{4},\frac{3\pi }{2},\pi \right\}$
So $\sum _{\theta \in S}{\sin }^2\left(\theta +\frac{\pi }{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$

Answer 2

Step 1: Analyze the condition

The given condition is: 

\[ \tan(\cos \theta) + \tan(\sin \theta) = 0. \] This implies: \[ \tan(\cos \theta) = -\tan(\sin \theta). \] Thus, we have the equation: \[ \cos \theta = - \sin \theta. \] From this, we can deduce: \[ \tan \theta = -1. \] 

Step 2: Solve for \( \theta \)

From \( \tan \theta = -1 \), the solutions in the interval \( [0, 2\pi) \) are:

\[ \theta = \frac{3\pi}{4}, \quad \theta = \frac{7\pi}{4}. \] 

Step 3: Evaluate \( \sin^2 \theta + \frac{\pi}{4} \)

We now compute the sum \( \sin^2 \theta + \frac{\pi}{4} \) for each solution of \( \theta \).

For \( \theta = \frac{3\pi}{4} \): \[ \sin^2 \left( \frac{3\pi}{4} \right) + \frac{\pi}{4} = \sin^2 \left( \pi \right) = 0. \] For \( \theta = \frac{7\pi}{4} \): \[ \sin^2 \left( \frac{7\pi}{4} \right) + \frac{\pi}{4} = \sin^2 \left( 2\pi \right) = 0. \] 

Conclusion

The sum is:

\[ \sum_{\theta \in S} \sin^2 \theta + \frac{\pi}{4} = 0 + 0 = 2. \]