Question 4 4.1. Given \( \sin \alpha=-\frac{8}{17} \), where \( 0^{\circ}<\alpha<270^{\circ} \), determine with the aid of a sketch, the value of the following: 4.1.1. \( \tan \alpha \) 4.1.2. \( \cos 2 \alpha \) 4.1.3. \( \sin \frac{\alpha}{2} \) 4.2. Simplify the following trigonometric expression, leaving your answer to only one trigonometric ratio: \( \left[\sin \left(x-1980^{\circ}\right)-\cos (-x)\right]^{2} \)
Upstudy AI Solution
Answer
Solution

Sign in to Unlock Answers for Free!
A Learning Platform Trusted by Millions of Real Students and Teachers.

Answered by UpStudy AI and reviewed by a Professional Tutor
Bonus Knowledge
To solve \( \tan \alpha \), we start with the fact that \( \sin \alpha = -\frac{8}{17} \). Since \( \alpha \) is in the third quadrant (where both sine and cosine are negative), we can find \( \cos \alpha \) using the Pythagorean identity: \( \sin^2 \alpha + \cos^2 \alpha = 1 \). This gives us \( \cos \alpha = -\sqrt{1 - \left(-\frac{8}{17}\right)^2} = -\frac{15}{17} \). Therefore, \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{8}{17}}{-\frac{15}{17}} = \frac{8}{15} \). Next, for \( \cos 2\alpha \), we use the double angle formula: \( \cos 2\alpha = 2\cos^2 \alpha - 1 \). Plugging in our value for \( \cos \alpha \), we find \( \cos 2\alpha = 2\left(-\frac{15}{17}\right)^2 - 1 = 2 \cdot \frac{225}{289} - 1 = \frac{450}{289} - 1 = \frac{450 - 289}{289} = \frac{161}{289} \). For \( \sin \frac{\alpha}{2} \), we use the half-angle formula \( \sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}} \). With \( \cos \alpha = -\frac{15}{17} \), we compute \( \sin \frac{\alpha}{2} = \sqrt{\frac{1 - (-\frac{15}{17})}{2}} = \sqrt{\frac{1 + \frac{15}{17}}{2}} = \sqrt{\frac{\frac{32}{17}}{2}} = \sqrt{\frac{32}{34}} = \sqrt{\frac{16}{17}} = \frac{4}{\sqrt{17}} \). Now for the expression \( \left[\sin \left(x-1980^{\circ}\right)-\cos (-x)\right]^{2} \) simplification: First, recall that \( \cos(-x) = \cos x \). Also, \( 1980^{\circ} \) can be simplified by subtracting \( 1800^{\circ} \) (which is equivalent to \( 5 \times 360^{\circ} \)): \( 1980^{\circ} = 180^{\circ} \). So we have \( \sin(x - 180^{\circ}) = -\sin x \). Therefore, the expression simplifies to \( \left[-\sin x - \cos x\right]^{2} = \left(-\sin x - \cos x\right)^{2} = (\sin x + \cos x)^{2} \). Thus, the final result is \( \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + \sin(2x) \).