Question

use the cosine of a sum and cosine of a difference identities to find \\( \cos (s+t) \\) and \\( \cos (s-t) \\).
\\( \sin s=\frac{5}{13} \\) and \\( \sin t=-\frac{4}{5} \\), s in quadrant ii and t in quadrant iv
\\( \cos (s+t)= \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
\\( \cos (s-t)= \\)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

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

Using the identity \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), for \(s\) with \(\sin s=\frac{5}{13}\) (in quadrant II where \(\cos s<0\)):

$$
LATEXBLOCK0
$$

For \(t\) with \(\sin t =-\frac{4}{5}\) (in quadrant IV where \(\cos t>0\)):

$$
LATEXBLOCK1
$$

Step2: Use the cosine of a sum 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 \(\cos s=-\frac{12}{13}\), \(\cos t=\frac{3}{5}\), \(\sin s=\frac{5}{13}\), \(\sin t =-\frac{4}{5}\)

$$
LATEXBLOCK2
$$

Step3: 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 \(\cos s=-\frac{12}{13}\), \(\cos t=\frac{3}{5}\), \(\sin s=\frac{5}{13}\), \(\sin t =-\frac{4}{5}\)

$$
LATEXBLOCK3
$$

Answer:

\(\cos(s + t)=-\frac{16}{65}\)
\(\cos(s - t)=-\frac{56}{65}\)