Question

without tables, evaluate $\sin(\alpha + \beta)$ and $\sin(\alpha - \beta)$. $\tan \alpha = \frac{4}{3}$, $\sec \beta = \frac{13}{12}$; $0 < \alpha < 90^\circ$, $0 < \beta < 90^\circ$

Explanation:

Step1: Find \(\sin\alpha\) and \(\cos\alpha\)

Given \(\tan\alpha=\frac{4}{3}=\frac{\text{opposite}}{\text{adjacent}}\) in a right - triangle. By the Pythagorean theorem, the hypotenuse \(r = \sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5\).
Since \(0\lt\alpha\lt90^{\circ}\), \(\sin\alpha=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{4}{5}\) and \(\cos\alpha=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{3}{5}\).

Step2: Find \(\sin\beta\) and \(\cos\beta\)

Given \(\sec\beta=\frac{13}{12}=\frac{1}{\cos\beta}\), so \(\cos\beta=\frac{12}{13}\).
Using the Pythagorean identity \(\sin^{2}\beta+\cos^{2}\beta = 1\), we have \(\sin\beta=\sqrt{1-\cos^{2}\beta}\).
Substitute \(\cos\beta=\frac{12}{13}\) into the formula: \(\sin\beta=\sqrt{1 - (\frac{12}{13})^{2}}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{169 - 144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}\) (since \(0\lt\beta\lt90^{\circ}\), \(\sin\beta\gt0\)).

Step3: Evaluate \(\sin(\alpha+\beta)\)

Use the sum formula for sine: \(\sin(A + B)=\sin A\cos B+\cos A\sin B\)
Substitute \(A=\alpha\) and \(B = \beta\):
\(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\)
\(=\frac{4}{5}\times\frac{12}{13}+\frac{3}{5}\times\frac{5}{13}\)
\(=\frac{48}{65}+\frac{15}{65}\)
\(=\frac{48 + 15}{65}=\frac{63}{65}\)

Step4: Evaluate \(\sin(\alpha - \beta)\)

Use the difference formula for sine: \(\sin(A - B)=\sin A\cos B-\cos A\sin B\)
Substitute \(A=\alpha\) and \(B=\beta\):
\(\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta\)
\(=\frac{4}{5}\times\frac{12}{13}-\frac{3}{5}\times\frac{5}{13}\)
\(=\frac{48}{65}-\frac{15}{65}\)
\(=\frac{48-15}{65}=\frac{33}{65}\)

Answer:

\(\sin(\alpha + \beta)=\frac{63}{65}\), \(\sin(\alpha-\beta)=\frac{33}{65}\)