## Trigonometry (11th Edition) Clone

$sin~2x = 0.6$ $cos~2x = -0.8$
Since $tan~x$ is positive and $sin~x$ is negative, the angle $x$ lies in quadrant III. $tan~x = \frac{opp}{adj} = 3 = \frac{-3}{-1}$ If the opposite side is -3 and the adjacent side is -1, then the hypotenuse is $\sqrt{(-3)^2+(-1)^2} = \sqrt{10}$ $sin~x = \frac{opp}{hyp} = \frac{-3~\sqrt{10}}{10}$ $cos~x = \frac{adj}{hyp} = \frac{-1~\sqrt{10}}{10}$ We can find $sin~2x$: $sin~2x = 2~sin~x~cos~x$ $sin~2x = (2)(\frac{-3~\sqrt{10}}{10})(\frac{-1~\sqrt{10}}{10})$ $sin~2x = \frac{60}{100}$ $sin~2x = 0.6$ We can find $cos~2x$: $cos~2x = cos^2~x-sin^2~x$ $cos~2x = (\frac{-1~\sqrt{10}}{10})^2-(\frac{-3~\sqrt{10}}{10})^2$ $cos~2x = \frac{10}{100}-\frac{90}{100}$ $cos~2x = \frac{-80}{100}$ $cos~2x = -0.8$