Unit Circle Functions
1. **Problem Statement:**
After one revolution, point A is at coordinates $$\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$ on the unit circle.
(i) Calculate all six trigonometric functions.
(ii) Determine the quadrant.
(iii) Calculate the angle formed by point A and its reference angle.
2. **Part (i): Calculate six trigonometric functions**
We know that for a point $$ (x, y) $$ on the unit circle at angle $$\theta$$,
- $$\sin \theta = y$$
- $$\cos \theta = x$$
- $$\tan \theta = \frac{y}{x}$$
- The reciprocal functions are:
- $$\csc \theta = \frac{1}{\sin \theta}$$
- $$\sec \theta = \frac{1}{\cos \theta}$$
- $$\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$$
Using the given coordinates:
- $$\sin \theta = \frac{1}{2}$$
- $$\cos \theta = -\frac{\sqrt{3}}{2}$$
- $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}$$
- $$\csc \theta = \frac{1}{\sin \theta} = 2$$
- $$\sec \theta = \frac{1}{\cos \theta} = -\frac{2}{\sqrt{3}}$$
- $$\cot \theta = \frac{1}{\tan \theta} = -\sqrt{3}$$
3. **Part (ii): Determine the quadrant**
- $$\cos \theta = x = -\frac{\sqrt{3}}{2} < 0$$ (Negative)
- $$\sin \theta = y = \frac{1}{2} > 0$$ (Positive)
- Cosine is negative, sine is positive, so point A lies in **Quadrant II**.
4. **Part (iii): Calculate the angle $$\theta$$ and reference angle**
- From $$ \sin \theta = \frac{1}{2} $$, and knowing the unit circle, possible angles are $$ 30^\circ $$ or $$ 150^\circ $$ (in degrees) or $$ \frac{\pi}{6} $$ or $$ \frac{5\pi}{6} $$ (in radians).
- Since point is in Quadrant II, $$\theta = 150^\circ = \frac{5\pi}{6}$$.
- The reference angle $$\alpha$$ is the acute angle with the x-axis:
$$\alpha = \pi - \theta = \pi - \frac{5\pi}{6} = \frac{\pi}{6} = 30^\circ$$.
**Final answers:**
(i) $$\sin \theta = \frac{1}{2}$$, $$\cos \theta = -\frac{\sqrt{3}}{2}$$, $$\tan \theta = -\frac{1}{\sqrt{3}}$$, $$\csc \theta = 2$$, $$\sec \theta = -\frac{2}{\sqrt{3}}$$, $$\cot \theta = -\sqrt{3}$$
(ii) Point A lies in **Quadrant II** because $$\cos \theta < 0$$ and $$\sin \theta > 0$$.
(iii) The angle $$\theta = \frac{5\pi}{6}$$ and the reference angle is $$\frac{\pi}{6}$$.