## Trigonometry (11th Edition) Clone

$$\cos x=-\frac{\sqrt{42}}{12}$$
$$\cos2x=-\frac{5}{12}\hspace{1.5cm}\frac{\pi}{2}\lt x\lt\pi\hspace{1.5cm}\cos x=?$$ Though this exercise can certainly be solved using half-angle identities, if you do not want to deal with the square root, another way to do it is through double-angle identities. - Double-angle Identity for cosines: $$\cos2x=2\cos^2x-1$$ $$\cos^2x=\frac{\cos2x+1}{2}$$ $$\cos^2x=\frac{-\frac{5}{12}+1}{2}$$ $$\cos^2x=\frac{\frac{7}{12}}{2}$$ $$\cos^2x=\frac{7}{24}$$ Thus, $$\cos x=\pm\frac{\sqrt7}{\sqrt{24}}=\pm\frac{\sqrt7}{2\sqrt6}=\pm\frac{\sqrt7\times\sqrt6}{2\times6}=\pm\frac{\sqrt{42}}{12}$$ As $\frac{\pi}{2}\lt x\lt \pi$, $x$ lies in the area of quadrant II, where cosines are negative. So $\cos x\lt0$. Therefore, $$\cos x=-\frac{\sqrt{42}}{12}$$