Respuesta :
Answer:
[tex]$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$[/tex]
Step-by-step explanation:
[tex]\alpha \text{ and } \beta \text{ in Quadrant I}[/tex]
[tex]$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10} $[/tex]
Using Pythagorean Identities:
[tex]\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}[/tex]
[tex]$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$[/tex]
[tex]\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}[/tex]
[tex]$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)} $[/tex]
[tex]$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$[/tex]
[tex]\text{Now let's find }\sin(\alpha)[/tex]
[tex]$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$[/tex]
The sum Identity is:
[tex]\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)[/tex]
I will just follow what the question asks.
[tex]\text{Find the value of }\alpha+2\beta[/tex]
[tex]\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)[/tex]
[tex]\text{I will first calculate }\cos(2\beta)[/tex]
[tex]$\cos(2\beta)=\frac{1-\tan^2(\beta)}{1+\tan^2(\beta)} =\frac{1-(\frac{1}{7})^2 }{1+(\frac{1}{7})^2}=\frac{24}{25}$[/tex]
[tex]\text{Now }\sin(2\beta)[/tex]
[tex]$\sin(2\beta)=2\sin(\beta)\cos(\beta)=2 \cdot \frac{\sqrt{10} }{10}\cdot \frac{3\sqrt{10} }{10} = \frac{3}{5} $[/tex]
Now we can perform the sum identity:
[tex]\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)[/tex]
[tex]$\sin(\alpha + 2\beta)=\frac{\sqrt{2}}{10}\cdot \frac{24}{25} +\frac{3}{5} \cdot \frac{7\sqrt{2} }{10} = \frac{129\sqrt{2}}{250}$[/tex]
But we are not done yet! You want
[tex]\alpha + 2\beta[/tex] and not [tex]\sin(\alpha + 2\beta)[/tex]
You actually want the
[tex]$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$[/tex]
