## Trigonometry (11th Edition) Clone

$$\cos(s-t)=\frac{4+6\sqrt6}{25}$$ $$\cos(s+t)=\frac{4-6\sqrt6}{25}$$
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 $\cos s$ and $\sin t$ are known, we would use Pythagorean Identities for $\sin$ and $\cos$ to find the rest: $$\sin^2 s=1-\cos^2 s=1-\Big(-\frac{1}{5}\Big)^2=1-\frac{1}{25}=\frac{24}{25}$$ $$\sin s=\pm\frac{\sqrt{24}}{5}$$ $$\cos^2 t=1-\sin^2 t=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$ $$\cos t=\pm\frac{4}{5}$$ Now about the signs of $\sin s$ and $\cos t$: As both $s$ and $t$ are in quadrant II, $\sin s\gt0$ while $\cos t\lt0$, so $$\sin s=\frac{\sqrt{24}}{5}\hspace{2cm}\cos t=-\frac{4}{5}$$ 2) Now we can apply cosine sum and difference identities: $$\cos(s+t)=\cos s\cos t-\sin s\sin t=\Big(-\frac{1}{5}\Big)\Big(-\frac{4}{5}\Big)-\frac{\sqrt{24}}{5}\frac{3}{5}$$ $$\cos(s+t)=\frac{4}{25}-\frac{3\sqrt{24}}{25}=\frac{4-3\sqrt{24}}{25}=\frac{4-6\sqrt6}{25}$$ $$\cos(s-t)=\cos s\cos t+\sin s\sin t=\Big(-\frac{1}{5}\Big)\Big(-\frac{4}{5}\Big)+\frac{\sqrt{24}}{5}\frac{3}{5}$$ $$\cos(s-t)=\frac{4}{25}+\frac{3\sqrt{24}}{25}=\frac{4+3\sqrt{24}}{25}=\frac{4+6\sqrt6}{25}$$