1. The problem asks to express trigonometric function values in terms of other trigonometric functions or non-trigonometric expressions given certain intervals and values. 2. We use the Pythagorean identities and definitions of tangent and cotangent: - $\sin^2 x + \cos^2 x = 1$ - $\tan x = \frac{\sin x}{\cos x}$ - $\cot x = \frac{\cos x}{\sin x}$ 3. For each case, we find the missing trigonometric values using the given information and the interval to determine the sign. **a)** $-\frac{\pi}{2} < x < 0$ and $\sin x = -0.6$ - Calculate $\cos x = \pm \sqrt{1 - \sin^2 x} = \pm \sqrt{1 - 0.36} = \pm 0.8$ - Since $x$ is in the fourth quadrant (angles between $-\frac{\pi}{2}$ and $0$), $\cos x > 0$, so $\cos x = 0.8$ **b)** $180^{\circ} < x < 270^{\circ}$ and $\sin x = -\frac{5}{13}$ - Calculate $\cos x = \pm \sqrt{1 - \sin^2 x} = \pm \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \pm \frac{12}{13}$ - In the third quadrant, $\cos x < 0$, so $\cos x = -\frac{12}{13}$ **c)** $\frac{3\pi}{2} < x < 2\pi$ and $\cos x = \frac{15}{17}$ - Calculate $\sin x = \pm \sqrt{1 - \cos^2 x} = \pm \sqrt{1 - \left(\frac{15}{17}\right)^2} = \pm \frac{8}{17}$ - In the fourth quadrant, $\sin x < 0$, so $\sin x = -\frac{8}{17}$ **d)** $90^{\circ} < x < 180^{\circ}$ and $\tan x = -\frac{3}{4}$ - Calculate $\sin x$ and $\cos x$ from $\tan x = \frac{\sin x}{\cos x}$ - Let $\sin x = 3k$, $\cos x = -4k$ (since $\tan x$ is negative and in the second quadrant $\sin x > 0$, $\cos x < 0$) - Use $\sin^2 x + \cos^2 x = 1$: $$ (3k)^2 + (-4k)^2 = 1 \Rightarrow 9k^2 + 16k^2 = 1 \Rightarrow 25k^2 = 1 \Rightarrow k = \frac{1}{5} $$ - So $\sin x = \frac{3}{5}$, $\cos x = -\frac{4}{5}$ **e)** $2 < x < 2\pi$ and $\cot x = -\frac{5}{12}$ - $\cot x = \frac{\cos x}{\sin x} = -\frac{5}{12}$ - Let $\cos x = -5k$, $\sin x = 12k$ - Use $\sin^2 x + \cos^2 x = 1$: $$ (-5k)^2 + (12k)^2 = 1 \Rightarrow 25k^2 + 144k^2 = 1 \Rightarrow 169k^2 = 1 \Rightarrow k = \frac{1}{13} $$ - So $\cos x = -\frac{5}{13}$, $\sin x = \frac{12}{13}$ **f)** $180^{\circ} < x < 270^{\circ}$ and $\cot x = \frac{12}{5}$ - $\cot x = \frac{\cos x}{\sin x} = \frac{12}{5}$ - Let $\cos x = 12k$, $\sin x = 5k$ - Use $\sin^2 x + \cos^2 x = 1$: $$ (12k)^2 + (5k)^2 = 1 \Rightarrow 144k^2 + 25k^2 = 1 \Rightarrow 169k^2 = 1 \Rightarrow k = \frac{1}{13} $$ - So $\cos x = \frac{12}{13}$, $\sin x = \frac{5}{13}$ - Since $x$ is in the third quadrant, both $\sin x$ and $\cos x$ are negative: $$ \cos x = -\frac{12}{13}, \quad \sin x = -\frac{5}{13} $$ Final answers: - a) $\cos x = 0.8$ - b) $\cos x = -\frac{12}{13}$ - c) $\sin x = -\frac{8}{17}$ - d) $\sin x = \frac{3}{5}$, $\cos x = -\frac{4}{5}$ - e) $\cos x = -\frac{5}{13}$, $\sin x = \frac{12}{13}$ - f) $\cos x = -\frac{12}{13}$, $\sin x = -\frac{5}{13}$