Trigonometry (11th Edition) Clone

$$\cos(s+t)=\frac{2\sqrt{638}-\sqrt{30}}{56}$$ $$\cos(s-t)=\frac{2\sqrt{638}+\sqrt{30}}{56}$$
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: $$\cos^2 s=1-\sin^2 s=1-\Big(\frac{\sqrt5}{7}\Big)^2=1-\frac{5}{49}=\frac{44}{49}$$ $$\cos s=\pm\frac{\sqrt{44}}{7}=\pm\frac{2\sqrt{11}}{7}$$ $$\cos^2 t=1-\sin^2 t=1-\Big(\frac{\sqrt6}{8}\Big)^2=1-\frac{6}{64}=\frac{58}{64}$$ $$\cos t=\pm\frac{\sqrt{58}}{8}$$ Now about the signs of $\cos s$ and $\cos t$: As both $s$ and $t$ are in quadrant I, $\cos s\gt0$ and also $\cos t\gt0$, so $$\cos s=\frac{2\sqrt{11}}{7}\hspace{2cm}\cos t=\frac{\sqrt{58}}{8}$$ 2) Now we can apply cosine sum and difference identities: $$\cos(s+t)=\cos s\cos t-\sin s\sin t=\Big(\frac{2\sqrt{11}}{7}\Big)\Big(\frac{\sqrt{58}}{8}\Big)-\Big(\frac{\sqrt5}{7}\Big)\Big(\frac{\sqrt6}{8}\Big)$$ $$\cos(s+t)=\frac{2\sqrt{638}}{56}-\frac{\sqrt{30}}{56}=\frac{2\sqrt{638}-\sqrt{30}}{56}$$ $$\cos(s-t)=\cos s\cos t+\sin s\sin t=\Big(\frac{2\sqrt{11}}{7}\Big)\Big(\frac{\sqrt{58}}{8}\Big)+\Big(\frac{\sqrt5}{7}\Big)\Big(\frac{\sqrt6}{8}\Big)$$ $$\cos(s-t)=\frac{2\sqrt{638}}{56}+\frac{\sqrt{30}}{56}=\frac{2\sqrt{638}+\sqrt{30}}{56}$$