Trigonometry (11th Edition) Clone

$$\cos(s+t)=\frac{2-2\sqrt{10}}{9}$$ $$\cos(s-t)=\frac{-2-2\sqrt{10}}{9}$$
To find $\cos(s+t)$ and $\cos(s-t)$, cosine sum and difference identities need to be applied. Yet, both cosine sum and difference identities rely on the fact that you already know $\cos s$, $\sin s$, $\cos t$ and $\sin t$. Therefore, the first job in this type of exercise is to find out all 4 values: $\cos s$, $\sin s$, $\cos t$ and $\sin t$. 1) Find $\cos s$, $\sin s$, $\cos t$ and $\sin t$. As $\sin s$ and $\sin t$ are known, we would use Pythagorean Identities for $\sin$ and $\cos$ to find the rest: $$\cos^2 s=1-\sin^2 s=1-\Big(\frac{2}{3}\Big)^2=1-\frac{4}{9}=\frac{5}{9}$$ $$\cos s=\pm\frac{\sqrt{5}}{3}$$ $$\cos^2 t=1-\sin^2 t=1-\Big(-\frac{1}{3}\Big)^2=1-\frac{1}{9}=\frac{8}{9}$$ $$\cos t=\pm\frac{\sqrt8}{3}=\pm\frac{2\sqrt2}{3}$$ Now about the signs of $\cos s$ and $\cos t$: As $s$ is in quadrant II, $\cos s\lt0$. At the same time, as $t$ is in quadrant IV, $\cos t\gt0$, so $$\cos s=-\frac{\sqrt{5}}{3}\hspace{2cm}\cos t=\frac{2\sqrt2}{3}$$ 2) Now we can apply cosine sum and difference identities: $$\cos(s+t)=\cos s\cos t-\sin s\sin t=\Big(-\frac{\sqrt5}{3}\Big)\Big(\frac{2\sqrt2}{3}\Big)-\frac{2}{3}\Big(-\frac{1}{3}\Big)$$ $$\cos(s+t)=-\frac{2\sqrt{10}}{9}+\frac{2}{9}=\frac{2-2\sqrt{10}}{9}$$ $$\cos(s-t)=\cos s\cos t+\sin s\sin t=\Big(-\frac{\sqrt5}{3}\Big)\Big(\frac{2\sqrt2}{3}\Big)+\frac{2}{3}\Big(-\frac{1}{3}\Big)$$ $$\cos(s-t)=-\frac{2\sqrt{10}}{9}-\frac{2}{9}=\frac{-2-2\sqrt{10}}{9}$$