Subjects trigonometry

Unit Circle Functions

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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 $$\frac{\pi}{6}$$.