## Trigonometry (11th Edition) Clone

$$\cos(s+t)=\frac{16}{65}$$ $$\cos(s-t)=-\frac{56}{65}$$
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 must be 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 cosine ones: $$\cos^2 s=1-\sin^2 s=1-\Big(\frac{3}{5}\Big)^2=1-\frac{9}{25}=\frac{16}{25}$$ $$\cos s=\pm\frac{4}{5}$$ $$\cos^2 t=1-\sin^2 t=1-\Big(-\frac{12}{13}\Big)^2=1-\frac{144}{169}=\frac{25}{169}$$ $$\cos t=\pm\frac{5}{13}$$ Now about the signs of $\cos s$ and $\cos t$: As $s$ is in quadrant I, $\cos s\gt0$, so $$\cos s=\frac{4}{5}$$ As $t$ is in quadrant III, $\cos t\lt0$, so $$\cos t=-\frac{5}{13}$$ 2) Now we apply cosine sum and difference identities: $$\cos(s+t)=\cos s\cos t-\sin s\sin t=\frac{4}{5}\Big(-\frac{5}{13}\Big)-\frac{3}{5}\Big(-\frac{12}{13}\Big)$$ $$\cos(s+t)=-\frac{4}{13}+\frac{36}{65}=\frac{16}{65}$$ $$\cos(s-t)=\cos s\cos t+\sin s\sin t=\frac{4}{5}\Big(-\frac{5}{13}\Big)+\frac{3}{5}\Big(-\frac{12}{13}\Big)$$ $$\cos(s-t)=-\frac{4}{13}-\frac{36}{65}=-\frac{56}{65}$$