Question

find the exact value of each of the following under the given conditions below.
\\(\tan\alpha = -\frac{3}{4},\frac{\pi}{2}<\alpha<\pi;\cos\beta = \frac{1}{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)=\\) (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(b) \\(\cos(\alpha + \beta)=\\) (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(c) \\(\sin(\alpha - \beta)=\\) (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)
(d) \\(\tan(\alpha - \beta)=\\) (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Find $\sin\alpha$ and $\cos\alpha$

Given $\tan\alpha =-\frac{3}{4}=\frac{\sin\alpha}{\cos\alpha}$ and $\sin^{2}\alpha+\cos^{2}\alpha = 1$. Since $\frac{\pi}{2}<\alpha<\pi$, $\sin\alpha>0$ and $\cos\alpha<0$. We have $\sin\alpha=\frac{3}{5}$ and $\cos\alpha =-\frac{4}{5}$.

Step2: Find $\sin\beta$

Given $\cos\beta=\frac{1}{2}$ and $0 <\beta<\frac{\pi}{2}$, using $\sin^{2}\beta+\cos^{2}\beta = 1$, we get $\sin\beta=\sqrt{1 - \cos^{2}\beta}=\frac{\sqrt{3}}{2}$.

Step3: Calculate $\sin(\alpha+\beta)$

Using the formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$, we have $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{3}{5}\times\frac{1}{2}+(-\frac{4}{5})\times\frac{\sqrt{3}}{2}=\frac{3 - 4\sqrt{3}}{10}$.

Step4: Calculate $\cos(\alpha+\beta)$

Using the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$, we have $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=(-\frac{4}{5})\times\frac{1}{2}-\frac{3}{5}\times\frac{\sqrt{3}}{2}=\frac{- 4-3\sqrt{3}}{10}$.

Step5: Calculate $\sin(\alpha - \beta)$

Using the formula $\sin(A - B)=\sin A\cos B-\cos A\sin B$, we have $\sin(\alpha - \beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta=\frac{3}{5}\times\frac{1}{2}-(-\frac{4}{5})\times\frac{\sqrt{3}}{2}=\frac{3 + 4\sqrt{3}}{10}$.

Step6: Calculate $\tan(\alpha - \beta)$

First, $\tan(\alpha - \beta)=\frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)}$. We find $\cos(\alpha - \beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta=(-\frac{4}{5})\times\frac{1}{2}+\frac{3}{5}\times\frac{\sqrt{3}}{2}=\frac{-4 + 3\sqrt{3}}{10}$. Then $\tan(\alpha - \beta)=\frac{\frac{3 + 4\sqrt{3}}{10}}{\frac{-4+3\sqrt{3}}{10}}=\frac{3 + 4\sqrt{3}}{-4 + 3\sqrt{3}}=\frac{(3 + 4\sqrt{3})(3\sqrt{3}+4)}{(3\sqrt{3}-4)(3\sqrt{3}+4)}=\frac{9\sqrt{3}+12 + 36+16\sqrt{3}}{27 - 16}=\frac{48 + 25\sqrt{3}}{11}$.

Answer:

(a) $\frac{3 - 4\sqrt{3}}{10}$
(b) $\frac{-4 - 3\sqrt{3}}{10}$
(c) $\frac{3 + 4\sqrt{3}}{10}$
(d) $\frac{48 + 25\sqrt{3}}{11}$