Question

without tables, evaluate \\( \cos(\alpha + \beta) \\) and \\( \cos(\alpha - \beta) \\).

1. \\( \sin \alpha = \frac{3}{5} \\), \\( \cos \beta = \frac{5}{13} \\); \\( 0 < \alpha < \frac{\pi}{2} \\), \\( 0 < \beta < \frac{\pi}{2} \\)
2. \\( \tan \alpha = \frac{4}{3} \\), \\( \cos \beta = \frac{12}{13} \\); \\( 0 < \alpha < \frac{\pi}{2} \\), \\( 0 < \beta < \frac{\pi}{2} \\)
3. \\( \tan \alpha = \frac{3}{4} \\), \\( \cos \beta = \frac{5}{13} \\); \\( 0 < \alpha < \frac{\pi}{2} \\), \\( 0 < \beta < \frac{\pi}{2} \\)
4. \\( \cos \alpha = -\frac{3}{5} \\), \\( \tan \beta = -\frac{5}{12} \\); neither \\( \alpha \\) nor \\( \beta \\) in quadrant ii
5. \\( \sin \alpha = -\frac{7}{25} \\), \\( \cot \beta = -\frac{8}{15} \\); neither \\( \alpha \\) nor \\( \beta \\) in quadrant iv

trigonometric

Explanation:

Step1: Solve for \(\cos\alpha\) and \(\sin\beta\)

Given \(\sin\alpha=\frac{3}{5}\) and \(0 < \alpha<\frac{\pi}{2}\), use \(\sin^{2}\alpha+\cos^{2}\alpha = 1\).
\(\cos\alpha=\sqrt{1-\sin^{2}\alpha}=\sqrt{1 - (\frac{3}{5})^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\) (since \(\alpha\) is in QI, \(\cos\alpha>0\)).

Given \(\cos\beta=\frac{5}{13}\) and \(0 < \beta<\frac{\pi}{2}\), use \(\sin^{2}\beta+\cos^{2}\beta = 1\).
\(\sin\beta=\sqrt{1-\cos^{2}\beta}=\sqrt{1 - (\frac{5}{13})^{2}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\) (since \(\beta\) is in QI, \(\sin\beta>0\)).

Step2: Apply \(\cos(\alpha+\beta)\) formula

The formula is \(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\).
Substitute values:
\(\cos(\alpha+\beta)=\frac{4}{5}\times\frac{5}{13}-\frac{3}{5}\times\frac{12}{13}=\frac{20}{65}-\frac{36}{65}=\frac{20 - 36}{65}=-\frac{16}{65}\).

Step3: Apply \(\cos(\alpha-\beta)\) formula

The formula is \(\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta\).
Substitute values:
\(\cos(\alpha-\beta)=\frac{4}{5}\times\frac{5}{13}+\frac{3}{5}\times\frac{12}{13}=\frac{20}{65}+\frac{36}{65}=\frac{20 + 36}{65}=\frac{56}{65}\).

Answer:

For problem 9: \(\cos(\alpha+\beta)=-\frac{16}{65}\), \(\cos(\alpha-\beta)=\frac{56}{65}\)