Question

If alpha and beta are the angles in the first quadrant tan alpha = 1/7 and sin beta =1/ root 10 then usind the formula sin (A +B) = sin A. CosB + sina. CosB find the value of alpha + 2beta​

Answer 1

Answer:

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Step-by-step explanation:

$\alpha \text{ and } \beta \text{ in Quadrant I}$

$\tan(\alpha)=\frac{1}{7} \text{ and } \sin(\beta)=\frac{1}{\sqrt{10}}=\frac{\sqrt{10} }{10}$

Using Pythagorean Identities:

$\boxed{\sin^2(\theta)+\cos^2(\theta)=1} \text{ and } \boxed{1+\tan^2(\theta)=\sec^2(\theta)}$

$\left(\frac{\sqrt{10} }{10} \right)^2+\cos^2(\beta)=1 \Longrightarrow \cos(\beta)=\sqrt{1-\frac{10}{100}} =\sqrt{\frac{90}{100}}=\frac{3\sqrt{10}}{10}$

$\text{Note: } \cos(\beta) \text{ is positive because the angle is in the first qudrant}$

$1+\left(\frac{1 }{7} \right)^2=\frac{1}{\cos^2(\alpha)} \Longrightarrow 1+\frac{1}{49}=\frac{1}{\cos^2(\alpha)} \Longrightarrow \frac{50}{49} =\frac{1}{\cos^2(\alpha)}$

$\Longrightarrow \frac{49}{50}=\cos^2(\alpha) \Longrightarrow \cos(\alpha)=\sqrt{\frac{49}{50} } =\frac{7\sqrt{2}}{10}$

$\text{Now let's find }\sin(\alpha)$

$\sin^2(\alpha)+\left(\frac{7\sqrt{2} }{10}\right)^2=1 \Longrightarrow \sin^2(\alpha) +\frac{49}{50}=1 \Longrightarrow \sin(\alpha)=\sqrt{1-\frac{49}{50}} = \frac{\sqrt{2}}{10}$

The sum Identity is:

$\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$

I will just follow what the question asks.

$\text{Find the value of }\alpha+2\beta$

$\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)$

$\text{I will first calculate }\cos(2\beta)$

$\cos(2\beta)=\frac{1-\tan^2(\beta)}{1+\tan^2(\beta)} =\frac{1-(\frac{1}{7})^2 }{1+(\frac{1}{7})^2}=\frac{24}{25}$

$\text{Now }\sin(2\beta)$

$\sin(2\beta)=2\sin(\beta)\cos(\beta)=2 \cdot \frac{\sqrt{10} }{10}\cdot \frac{3\sqrt{10} }{10} = \frac{3}{5}$

Now we can perform the sum identity:

$\sin(\alpha + 2\beta)=\sin(\alpha)\cos(2\beta)+\sin(2\beta)\cos(\alpha)$

$\sin(\alpha + 2\beta)=\frac{\sqrt{2}}{10}\cdot \frac{24}{25} +\frac{3}{5} \cdot \frac{7\sqrt{2} }{10} = \frac{129\sqrt{2}}{250}$

But we are not done yet! You want

$\alpha + 2\beta$ and not $\sin(\alpha + 2\beta)$

You actually want the

$\arcsin\left(\frac{129\sqrt{2}}{250}\right)\approx 0.8179$

Answer 2

Answer:

ok bye guy................