## Trigonometry (11th Edition) Clone

$$\cos\theta=-\frac{\sqrt3}{2}$$
$$\cos2\theta=\frac{1}{2}\hspace{1cm}\theta\hspace{0.2cm}\text{terminates in quadrant II}\hspace{1cm}\cos\theta=?$$ From the original half-angle identity for cosines: $$\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$ we can rewrite to use to find $\cos\theta$ $$\cos\theta=\pm\sqrt{\frac{1+\cos2\theta}{2}}$$ Also, whether to choose positive or negative square root now depends on the sign of $\cos\theta$. $\theta$ terminates in quadrant II, where cosines are negative. Therefore, $\cos\theta\lt0$. As a result, we need to choose negative square root: $$\cos\theta=-\sqrt{\frac{1+\cos2\theta}{2}}$$ Now we can start calculating $\cos\theta$. $$\cos\theta=-\sqrt{\frac{1+\frac{1}{2}}{2}}$$ $$\cos\theta=-\sqrt{\frac{\frac{3}{2}}{2}}$$ $$\cos\theta=-\sqrt{\frac{3}{4}}$$ $$\cos\theta=-\frac{\sqrt3}{2}$$ Therefore, $$\cos\theta=-\frac{\sqrt3}{2}$$