find the exact value of each of the following under the given conditions. a. \\( \\cos (\\alpha+\\beta) \\)…

Explanation:

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

Given $\tan\alpha=\frac{1}{2},\pi<\alpha<\frac{3\pi}{2}$. Since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{1}{2}$, so $\cos\alpha = 2\sin\alpha$. Using $\sin^{2}\alpha+\cos^{2}\alpha = 1$, substitute $\cos\alpha = 2\sin\alpha$: $$\sin^{2}\alpha+(2\sin\alpha)^{2}=1$$ $$\sin^{2}\alpha + 4\sin^{2}\alpha=1$$ $$5\sin^{2}\alpha=1$$ $$\sin\alpha=-\frac{1}{\sqrt{5}}=-\frac{\sqrt{5}}{5}$$ (because $\pi<\alpha<\frac{3\pi}{2}$, $\sin\alpha<0$) Then $\cos\alpha=-\frac{2\sqrt{5}}{5}$ (because $\pi<\alpha<\frac{3\pi}{2}$, $\cos\alpha<0$)

Step2: Find $\sin\beta$

Given $\cos\beta=\frac{3}{5},\frac{3\pi}{2}<\beta<2\pi$. Using $\sin^{2}\beta+\cos^{2}\beta = 1$: $$\sin\beta=-\sqrt{1 - \cos^{2}\beta}=-\sqrt{1-\left(\frac{3}{5}\right)^{2}}=-\frac{4}{5}$$ (because $\frac{3\pi}{2}<\beta<2\pi$, $\sin\beta<0$)

Step3: Find $\cos(\alpha+\beta)$

Using the formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$: $$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$$ Substitute $\cos\alpha=-\frac{2\sqrt{5}}{5},\sin\alpha=-\frac{\sqrt{5}}{5},\cos\beta=\frac{3}{5},\sin\beta=-\frac{4}{5}$: $$\cos(\alpha + \beta)=\left(-\frac{2\sqrt{5}}{5}\right)\times\frac{3}{5}-\left(-\frac{\sqrt{5}}{5}\right)\times\left(-\frac{4}{5}\right)$$ $$=\frac{-6\sqrt{5}-4\sqrt{5}}{25}=-\frac{10\sqrt{5}}{25}=-\frac{2\sqrt{5}}{5}$$

Step4: Find $\sin(\alpha+\beta)$

Using the formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$: $$\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta$$ Substitute the values: $$\sin(\alpha+\beta)=\left(-\frac{\sqrt{5}}{5}\right)\times\frac{3}{5}+\left(-\frac{2\sqrt{5}}{5}\right)\times\left(-\frac{4}{5}\right)$$ $$=\frac{-3\sqrt{5}+8\sqrt{5}}{25}=\frac{5\sqrt{5}}{25}=\frac{\sqrt{5}}{5}$$

Step5: Find $\tan(\alpha+\beta)$

Using the formula $\tan(A + B)=\frac{\sin(A + B)}{\cos(A + B)}$: $$\tan(\alpha+\beta)=\frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}}=-\frac{1}{2}$$

Answer:

a. $\cos(\alpha+\beta)=-\frac{2\sqrt{5}}{5}$

b. $\sin(\alpha+\beta)=\frac{\sqrt{5}}{5}$

c. $\tan(\alpha+\beta)=-\frac{1}{2}$