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

$\tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0$
$ \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
$-\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2}$
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 $\displaystyle\sum_{\theta \in S} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$