use the cosine of a sum and cosine of a difference identities to find $ \cos (s+t) $ and $ \cos (s-t) $.
$ \sin s=-\frac{3}{5} $ and $ \sin t=\frac{5}{13}, s $ in quadrant iii and $ t $ in quadrant i
$ \cos (s+t)=-\frac{33}{65} $
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
$ \cos (s-t)=\square $
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

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Accepted Answer

# Explanation:
## Step1: Find $\cos s$ and $\cos t$
Using the Pythagorean identity $\sin^{2}\alpha+\cos^{2}\alpha = 1$, for $s$ with $\sin s=-\frac{3}{5}$ (in quadrant III, so $\cos s<0$):
$$
\begin{align*}
\cos^{2}s&=1-\sin^{2}s\
\cos^{2}s&=1 - (-\frac{3}{5})^{2}\
\cos^{2}s&=1-\frac{9}{25}\
\cos^{2}s&=\frac{16}{25}\
\cos s&=-\frac{4}{5}
\end{align*}
$$
For $t$ with $\sin t=\frac{5}{13}$ (in quadrant I, so $\cos t>0$):
$$
\begin{align*}
\cos^{2}t&=1-\sin^{2}t\
\cos^{2}t&=1-(\frac{5}{13})^{2}\
\cos^{2}t&=1 - \frac{25}{169}\
\cos^{2}t&=\frac{144}{169}\
\cos t&=\frac{12}{13}
\end{align*}
$$
## Step2: Use the cosine of a difference identity $\cos(A - B)=\cos A\cos B+\sin A\sin B$
Here $A = s$ and $B=t$, so $\cos(s - t)=\cos s\cos t+\sin s\sin t$
Substitute $\cos s=-\frac{4}{5}$, $\cos t=\frac{12}{13}$, $\sin s=-\frac{3}{5}$, $\sin t=\frac{5}{13}$
$$
\begin{align*}
\cos(s - t)&=(-\frac{4}{5})\times\frac{12}{13}+(-\frac{3}{5})\times\frac{5}{13}\
&=-\frac{48}{65}-\frac{15}{65}\
&=-\frac{63}{65}
\end{align*}
$$

# Answer:
$-\frac{63}{65}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find \(\cos s\) and \(\cos t\)

Step2: Use the cosine of a difference identity \(\cos(A - B)=\cos A\cos B+\sin A\sin B\)

Answer: