Question

find the exact value of each of the following under the given conditions.
tan α = -\frac{8}{15}, α lies in quadrant ii, and cos β = \frac{5}{8}, β lies in quadrant i
a. sin (α + β) b. cos (α + β) c. tan (α + β)
(simplify your answer. type an exact answer, using radicals as needed. rationalize all denominators. use integers or
fractions for any numbers in the expression.)
c. tan (α + β) = \square
(simplify your answer. type an exact answer, using radicals as needed. rationalize all denominators. use integers or
fractions for any numbers in the expression.)

Explanation:

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

• Given \(\tan\alpha =-\frac{8}{15}\) and \(\alpha\) in quadrant II. Using \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\) and \(\sin^{2}\alpha+\cos^{2}\alpha = 1\), we have \(\sin\alpha=\frac{8}{17}\), \(\cos\alpha=-\frac{15}{17}\).
• Given \(\cos\beta=\frac{5}{8}\) and \(\beta\) in quadrant I. Using \(\sin^{2}\beta+\cos^{2}\beta = 1\), we get \(\sin\beta=\frac{\sqrt{39}}{8}\).

Step2: Use the formula \(\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\)

• First find \(\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\sqrt{39}}{5}\)
• Then \(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1 - \tan\alpha\tan\beta}=\frac{-\frac{8}{15}+\frac{\sqrt{39}}{5}}{1-(-\frac{8}{15})\times\frac{\sqrt{39}}{5}}=\frac{- 8 + 3\sqrt{39}}{15 + \frac{8\sqrt{39}}{5}}=\frac{(-8 + 3\sqrt{39})\times5}{75+8\sqrt{39}}=\frac{-40+15\sqrt{39}}{75 + 8\sqrt{39}}\times\frac{75-8\sqrt{39}}{75-8\sqrt{39}}\)

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Answer:

\(\frac{1445\sqrt{39}-7680}{3129}\)