+118723 cos2theta = -4/5 and 90* < theta < 135*, what is the value of costheta
\(cos2\theta = \frac{-4}{5}\\ cos^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-sin^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-2sin^2\theta = \frac{-4}{5}\\ 1+\frac{4}{5} = 2sin^2\theta\\ \frac{9}{5} = 2sin^2\theta\\ \frac{9}{10} = sin^2\theta\\ sin\theta=\pm\frac{3}{\sqrt{10}} \\ \mbox{But theta is in the second quadrant}\\ sin\theta=\frac{3}{\sqrt{10}} \\ \theta = 180-sin^{-1}\frac{3}{\sqrt{10}} \\ \theta \approx 108^026'58" \)
+118723 cos2theta = -4/5 and 90* < theta < 135*, what is the value of costheta
\(cos2\theta = \frac{-4}{5}\\ cos^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-sin^2\theta - sin^2\theta = \frac{-4}{5}\\ 1-2sin^2\theta = \frac{-4}{5}\\ 1+\frac{4}{5} = 2sin^2\theta\\ \frac{9}{5} = 2sin^2\theta\\ \frac{9}{10} = sin^2\theta\\ sin\theta=\pm\frac{3}{\sqrt{10}} \\ \mbox{But theta is in the second quadrant}\\ sin\theta=\frac{3}{\sqrt{10}} \\ \theta = 180-sin^{-1}\frac{3}{\sqrt{10}} \\ \theta \approx 108^026'58" \)
+37170 Cos 2theta = -4/5
ARCcos2 theta = arcos (-4/5)
2theta = 143.13 degrees
theta = 71.565 degrees BUT this is in the WRONG range of 90 - 135 SO we need to find the value in the second quandrant which corresponds to 71.565
180 - 71.565 = 108.435 degrees
