Trigonometric Ratios Question 1
Question 1 - 24 January - Shift 2
Let $S={\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0}$. Then $\sum _{\theta \in S} \sin ^{2}(\theta+\frac{\pi}{4})$ is equal to ______
Show Answer
Answer: 2
Solution:
Formula: General solution of Trigonometric Equations, Even and Odd Formulas, Range of Trigonometric Expression
$\tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0$
$\tan (\pi \cos \theta)=-\tan (\pi \sin \theta)$
$\tan (\pi \cos \theta)=\tan (-\pi \sin \theta)$
$\pi \cos \theta=n \pi-\pi \sin \theta$
$\sin \theta+\cos \theta=n$ 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 $\theta \in{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi}$
So $\sum _{\theta \in S} \sin ^{2}(\theta+\frac{\pi}{4})=2(0)+4(\frac{1}{2})=2$
