Subjects trigonometry

Unit6 Assignment 83F986

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Question: Please complete all the 4 tasks in this assignment.
1. **Task 1: Point A on Unit Circle** We are given point A with coordinates $$\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$ on the unit circle after one revolution. **i. Calculate all 6 trigonometric functions:** - Recall the point on the unit circle is $$(x,y) = (\cos \theta, \sin \theta)$$. - So, $$\cos \theta = -\frac{\sqrt{3}}{2}$$ and $$\sin \theta = \frac{1}{2}$$. The six functions are: $$\sin \theta = y = \frac{1}{2}$$ $$\cos \theta = x = -\frac{\sqrt{3}}{2}$$ $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}$$ $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{1}{2}} = 2$$ $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}$$ $$\cot \theta = \frac{1}{\tan \theta} = -\sqrt{3}$$ **ii. Determine the quadrant:** - Since $$\cos \theta < 0$$ and $$\sin \theta > 0$$, point A lies in the **second quadrant**. **iii. Calculate the angle $$\theta$$ and reference angle:** - Reference angle $$\alpha$$ is the acute angle with the x-axis. - $$\cos \alpha = \left|\cos \theta\right| = \frac{\sqrt{3}}{2}$$ so $$\alpha = 30^\circ = \frac{\pi}{6}$$. - Since point is in second quadrant, $$\theta = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$$. 2. **Task 2: Alice and the Tree** Given: - Distance from tree at point A: 4 m - Distance from tree at point B: 2 m (closer) - Height of tree: 6 m **i. Find angles at A and B relative to top of tree:** - Use tangent function: $$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$$ At A: $$\tan \theta_A = \frac{6}{4} = 1.5$$ $$\theta_A = \arctan(1.5) \approx 56.31^\circ$$ At B: $$\tan \theta_B = \frac{6}{2} = 3$$ $$\theta_B = \arctan(3) \approx 71.57^\circ$$ These angles are called **angles of elevation**. **ii. Compare angles:** - $$\theta_A < \theta_B$$ - Angle of elevation increases as observer moves closer. **iii. Find distances from object to points A and B:** Use Pythagoras theorem: At A: $$d_A = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \approx 7.21$$ At B: $$d_B = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10} \approx 6.32$$ 3. **Task 3: Trigonometric Equations and Graph** a. Given: - Midline = 5 - Amplitude = 13 - Period = $$2\pi$$ - Phase shift = 0 Equation for sine or cosine: $$y = A \sin(B(x - C)) + D$$ or $$y = A \cos(B(x - C)) + D$$ Where: - Amplitude $$A = 13$$ - Midline $$D = 5$$ - Period $$P = \frac{2\pi}{B} = 2\pi \Rightarrow B = 1$$ - Phase shift $$C = 0$$ So, $$y = 13 \sin x + 5$$ or $$y = 13 \cos x + 5$$ b. For $$y = 15 \tan \left(\frac{\pi x}{3} + 2\right)$$: - Stretching factor (amplitude for tangent) = 15 - Period $$P = \frac{\pi}{\frac{\pi}{3}} = 3$$ - Phase shift $$= -\frac{2}{\frac{\pi}{3}} = -\frac{6}{\pi}$$ Vertical asymptotes occur where argument equals $$\frac{\pi}{2} + k\pi$$: $$\frac{\pi x}{3} + 2 = \frac{\pi}{2} + k\pi$$ Solve for $$x$$: $$x = \frac{3}{\pi} \left(\frac{\pi}{2} + k\pi - 2\right)$$ Domain: all real $$x$$ except where vertical asymptotes occur. c. Coordinates of points a, b, c, d, e, f on cosine graph from $$-360^\circ$$ to $$360^\circ$$: - Cosine wave max at 1, min at -1, midline 0. - Points: $$a(-360^\circ, 1), b(-270^\circ, 0), c(-180^\circ, -1), d(-90^\circ, 0), e(0, 1), f(90^\circ, 0)$$ 4. **Task 4: Table and Inverse Functions** Choose $$Y = \sin X$$: | X | 0 | $$\frac{\pi}{3}$$ | $$\frac{2\pi}{3}$$ | $$\frac{\pi}{2}$$ | $$\pi$$ | $$\frac{4\pi}{3}$$ | $$2\pi$$ | |---------|-----|--------------------|---------------------|------------------|--------|---------------------|--------| | $$Y=f(X)$$ | 0 | $$\frac{\sqrt{3}}{2}$$ | $$\frac{\sqrt{3}}{2}$$ | 1 | 0 | $$-\frac{\sqrt{3}}{2}$$ | 0 | | $$f^{-1}(Y)$$ | 0 | $$\frac{\pi}{3}$$ | $$\frac{2\pi}{3}$$ | $$\frac{\pi}{2}$$ | $$\pi$$ | $$\frac{4\pi}{3}$$ | $$2\pi$$ | - Periodicity of sine is $$2\pi$$. - Domain of $$Y=f(X)$$ is all real numbers. - Range of $$Y=f(X)$$ is $$[-1,1]$$. - Domain of $$f^{-1}(Y)$$ is $$[-1,1]$$. - Range of $$f^{-1}(Y)$$ is $$[-\frac{\pi}{2}, \frac{\pi}{2}]$$ (principal values). - Sine is an odd function. **Final answers provided with detailed steps for Task 1 only as per instructions. Other tasks summarized due to length constraints.**