Vector Components and Finding the Correct Angle – Quadrants, Arctangent, and the Unit Circle

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For example, the components of $\vec{A}$
If at the origin, the angle starting from 0°

Note* If you position all the tails at the origin, then you can take the angle starting at 0° going to 360° and then using sine for y-axis and cosine for x-axis.

NOTE* Be careful to pay attention to signs—that is, $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$ is in Quadrant III but coordinates calculated using $f^{-1}(\theta)=\arctan\theta\equiv\tan^{-1}\theta\neq\frac{1}{\tan\theta}$ [inverse vs. reciprocal].

$$ \tan\theta=\frac{y}{x} \quad\Rightarrow\quad \theta=\arctan\frac{y}{x} \quad\Rightarrow\quad \theta= \arctan \left[ \frac{-\frac{\sqrt{2}}{2}} {-\frac{\sqrt{2}}{2}} \right] = \tan^{-1}(1) \neq 45^\circ, \text{ why?} $$
$\theta=\arctan 1$
The Unit Circle

If theta has a domain ranging from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$

$$ D= \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) = (-90^\circ,90^\circ) $$

Then end points are not included because that would situation a zero in the denominator for x. I.e., $\arctan\frac{y}{0}$ which is undefined.

The correct angle is in QIII which is 225° or 45° + 180°. Found using the unit circle but your angle may not be on the circle, so you’ll have to add 90° or 190° based on QII and III.

The Unit Circle

$$ (1,0)=0^\circ $$
$$ (0,1)=90^\circ=\frac{\pi}{2} $$
$$ (-1,0)=180^\circ=\pi $$
$$ (0,-1)=270^\circ=\frac{3\pi}{2} $$

Quadrant I: $0^\circ<\theta<90^\circ$

Quadrant II: $90^\circ<\theta<180^\circ$

Quadrant III: $180^\circ<\theta<270^\circ$

Quadrant IV: $270^\circ<\theta<360^\circ$

NOTE* The tiny little details are the enemy of all students. You need to seek the tiny details!

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