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 numbers in the expre
(b) (cos (alpha+\beta)=-\frac{12 sqrt{3}+5}{26})
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expre
(c) (sin (alpha-\beta)=\frac{5 sqrt{3}+12}{26})
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expre
(d) (\tan (alpha-\beta)=)
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expre

Explanation:

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

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

Step2: Find \(\sin\beta\)

Since \(\cos\beta=\frac{\sqrt{3}}{2}\) and \(\sin^{2}\beta+\cos^{2}\beta = 1\), \(0 <\beta<\frac{\pi}{2}\) (so \(\sin\beta>0\)). Then \(\sin\beta=\sqrt{1-\cos^{2}\beta}=\sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}=\frac{1}{2}\).

Step3: Use the formula \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\)

We know \(\tan\alpha=-\frac{5}{12}\), \(\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\).

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Answer:

\(-\frac{240 + 169\sqrt{3}}{463}\)