## Trigonometry (11th Edition) Clone

$sin(\theta) = \frac{\sqrt{14}}{4}$ $cos(\theta) = \frac{\sqrt{2}}{4}$
If $~~90^{\circ} \lt 2\theta \lt 180^{\circ},~~$ then $~~45^{\circ} \lt \theta \lt 90^{\circ},~~$ so $\theta$ is in quadrant I. Since $\theta$ is in quadrant I, $sin~\theta$ is positive. We can find the value of $sin~\theta$: $cos~2\theta = 1-2~sin^2~\theta$ $sin^2~\theta = \frac{1-cos~2\theta}{2}$ $sin`\theta = \sqrt{\frac{1-cos~2\theta}{2}}$ $sin~\theta = \sqrt{\frac{1-(-\frac{3}{4})}{2}}$ $sin~\theta = \sqrt{\frac{(\frac{7}{4})}{2}}$ $sin~\theta = \sqrt{\frac{7}{8}}$ $sin~\theta = \frac{\sqrt{7}}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}$ $sin~\theta = \frac{\sqrt{14}}{4}$ Since $\theta$ is in quadrant I, $cos~\theta$ is positive. We can find the value of $cos~\theta$: $cos~2\theta = 2~cos^2~\theta-1$ $cos^2~\theta = \frac{1+cos~2\theta}{2}$ $cos~\theta = \sqrt{\frac{1+cos~2\theta}{2}}$ $cos~\theta = \sqrt{\frac{1+(-\frac{3}{4})}{2}}$ $cos~\theta = \sqrt{\frac{(\frac{1}{4})}{2}}$ $cos~\theta = \sqrt{\frac{1}{8}}$ $cos~\theta = \frac{1}{2\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}$ $cos~\theta = \frac{\sqrt{2}}{4}$