Question

find the exact value of each of the following under the given conditions below.

\\( \tan \alpha=-\frac{5}{12}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{\sqrt{3}}{2}, 0<\beta<\frac{\pi}{2} \\)

(a) \\( \sin (\alpha+\beta) \\) (b) \\( \cos (\alpha+\beta) \\) (c) \\( \sin (\alpha-\beta) \\) (d) \\( \tan (\alpha-\beta) \\)

(a) \\( \sin (\alpha+\beta)=\frac{5 \sqrt{3}-12}{26} \\)
(simplify your answer, including any radicals. use integers or fractions for any numbe

(b) \\( \cos (\alpha+\beta)=\\)
(simplify your answer, including any radicals. use integers or fractions for any number

Explanation:

Step1: Find \(\sin\alpha\), \(\cos\alpha\), \(\sin\beta\)

• Given \(\tan\alpha =-\frac{5}{12}=\frac{\sin\alpha}{\cos\alpha}\) and \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), \(\frac{\pi}{2}<\alpha<\pi\).
• Let \(\sin\alpha = 5k\), \(\cos\alpha=- 12k\), then \((5k)^{2}+(-12k)^{2}=1\), \(25k^{2}+144k^{2}=1\), \(169k^{2}=1\), \(k=\frac{1}{13}\) (since \(\frac{\pi}{2}<\alpha<\pi\), \(\sin\alpha>0\), \(\cos\alpha<0\)). So \(\sin\alpha=\frac{5}{13}\), \(\cos\alpha =-\frac{12}{13}\).
• Given \(\cos\beta=\frac{\sqrt{3}}{2}\), \(0 <\beta<\frac{\pi}{2}\), then \(\sin\beta=\sqrt{1-\cos^{2}\beta}=\sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}=\frac{1}{2}\).

Step2: Use the formula \(\cos(A + B)=\cos A\cos B-\sin A\sin B\)

• Substitute \(A=\alpha\), \(B = \beta\), \(\cos\alpha=-\frac{12}{13}\), \(\sin\alpha=\frac{5}{13}\), \(\cos\beta=\frac{\sqrt{3}}{2}\), \(\sin\beta=\frac{1}{2}\) into the formula.
• \(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=(-\frac{12}{13})\times\frac{\sqrt{3}}{2}-\frac{5}{13}\times\frac{1}{2}=-\frac{12\sqrt{3}+5}{26}\)

Answer:

\(-\frac{12\sqrt{3}+5}{26}\)