Question

use the given information to find (a) sin (s + t), (b) tan (s + t), and (c) the quadrant of s + t. sin s = \frac{5}{7} and sin t = - \frac{6}{7}, s in quadrant ii and t in quadrant iv (a) sin (s + t) = \square (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find \(\cos s\)

Since \(\sin s=\frac{5}{7}\) and \(s\) is in quadrant II. Using \(\sin^{2}\theta+\cos^{2}\theta = 1\), we have \(\cos^{2}s=1-\sin^{2}s=1 - (\frac{5}{7})^{2}=1-\frac{25}{49}=\frac{24}{49}\). So \(\cos s=-\frac{2\sqrt{6}}{7}\) (negative in quadrant II).

Step2: Find \(\cos t\)

Since \(\sin t =-\frac{6}{7}\) and \(t\) is in quadrant IV. Using \(\sin^{2}\theta+\cos^{2}\theta = 1\), we have \(\cos^{2}t=1-\sin^{2}t=1-(-\frac{6}{7})^{2}=1-\frac{36}{49}=\frac{13}{49}\). So \(\cos t=\frac{\sqrt{13}}{7}\) (positive in quadrant IV).

Step3: Use the sum formula for sine

The sum formula for \(\sin(A + B)=\sin A\cos B+\cos A\sin B\). Here \(A = s\) and \(B=t\).
\(\sin(s + t)=\sin s\cos t+\cos s\sin t\)
Substitute \(\sin s=\frac{5}{7}\), \(\cos s=-\frac{2\sqrt{6}}{7}\), \(\sin t=-\frac{6}{7}\), \(\cos t=\frac{\sqrt{13}}{7}\)

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Answer:

\(\frac{5\sqrt{13}+12\sqrt{6}}{49}\)