If $\alpha$ is in the $3^{ ext {rd }}$ quadrant, $\beta$ is in the $2^{ ext {nd }}$ Quadrant such that $ an \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}}$, then $\sin (2 \alpha+\beta)$

Question

If $\alpha$ is in the $3^{	ext {rd }}$ quadrant, $\beta$ is in the $2^{	ext {nd }}$ Quadrant such that $	an \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}}$, then $\sin (2 \alpha+\beta)$

Aicharya · Accepted Answer

$	an \alpha=\frac{1}{7} \Rightarrow \cos 2 \alpha=\frac{1-	an ^2 \alpha}{1+	an ^2 \alpha}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25}$ $\sin 2 \alpha=\frac{2 	an \alpha}{1+	an ^2 \alpha}=\frac{\frac{2}{7}}{1+\frac{1}{49}}=\frac{14}{50}=\frac{7}{25}$ Also, $\sin \beta=\frac{1}{\sqrt{10}} \Rightarrow \cos \beta=\frac{-3}{\sqrt{10}}$ [ $\because \beta$ lies in 2nd quadrant] So, $\sin (2 \alpha+\beta)=\sin 2 \alpha \cos \beta+\cos 2 \alpha \sin \beta$ $\Rightarrow \sin (2 \alpha+\beta)=\frac{-21}{25 \sqrt{10}}+\frac{24}{25 \sqrt{10}}=\frac{3}{25 \sqrt{10}}$.

If $\alpha$ is in the $3^{\text {rd }}$ quadrant, $\beta$ is in the $2^{\text {nd }}$ Quadrant such that…

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