what is arctan(-1)?\n-\\frac{3\\pi}{4}\n-\\frac{\\pi}{4}\n\\frac{\\pi}{4}\n\\frac{3\\pi}{4}
what is arctan(-1)?\n-\\frac{3\\pi}{4}\n-\\frac{\\pi}{4}\n\\frac{\\pi}{4}\n\\frac{3\\pi}{4}
Answer
Answer:
B. $-\frac{\pi}{4}$
Explanation:
Step1: Recall arctan definition
The range of $y = \arctan(x)$ is $(-\frac{\pi}{2},\frac{\pi}{2})$. We need to find an angle $\theta$ in this range such that $\tan(\theta)=- 1$.
Step2: Use tangent - angle relationship
We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$, and $\tan(-\frac{\pi}{4})=\frac{\sin(-\frac{\pi}{4})}{\cos(-\frac{\pi}{4})}=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=-1$. Also, $-\frac{\pi}{4}\in(-\frac{\pi}{2},\frac{\pi}{2})$. So, $\arctan(-1)=-\frac{\pi}{4}$.