If alpha is in the 3^{text{rd}} quadrant,beta | vedclass

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(C) Given $	an \alpha = \frac{1}{7}$. Since $\alpha$ is in the $3^{	ext{rd}}$ quadrant,$\sin 2\alpha = \frac{2 	an \alpha}{1 + 	an^2 \alpha} = \fr

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$\frac{3}{25 \sqrt{10}}$

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(C) Given $	an \alpha = \frac{1}{7}$. Since $\alpha$ is in the $3^{	ext{rd}}$ quadrant,$\sin 2\alpha = \frac{2 	an \alpha}{1 + 	an^2 \alpha} = \frac{2/7}{1 + 1/49} = \frac{14/7}{50/49} = \frac{14}{50} = \frac{7}{25}$.$\cos 2\alpha = \frac{1 - 	an^2 \alpha}{1 + 	an^2 \alpha} = \frac{1 - 1/49}{1 + 1/49} = \frac{48/49}{50/49} = \frac{48}{50} = \frac{24}{25}$.Given $\sin \beta = \frac{1}{\sqrt{10}}$. Since $\beta$ is in the $2^{	ext{nd}}$ quadrant,$\cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - 1/10} = -\sqrt{9/10} = -\frac{3}{\sqrt{10}}$.Using the formula $\sin(2\alpha + \beta) = \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta$:$\sin(2\alpha + \beta) = \left(\frac{7}{25}ight)\left(-\frac{3}{\sqrt{10}}ight) + \left(\frac{24}{25}ight)\left(\frac{1}{\sqrt{10}}ight)$$= -\frac{21}{25\sqrt{10}} + \frac{24}{25\sqrt{10}} = \frac{3}{25\sqrt{10}}$.

If $\alpha$ is in the $3^{\text{rd}}$ quadrant,$\beta$ is in the $2^{\text{nd}}$ quadrant such that $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$,then find the value of $\sin(2\alpha + \beta)$.

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