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

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

Answer 2

`$\arcsin\left((129√(2))/(250)\right)\approx 0.8179$`

Explanation:

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

`$\tan(\alpha)=(1)/(7) \text{ and } \sin(\beta)=(1)/(√(10))=(√(10) )/(10) $`

Using Pythagorean Identities:

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

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

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

`$1+\left((1 )/(7) \right)^2=(1)/(\cos^2(\alpha)) \Longrightarrow 1+(1)/(49)=(1)/(\cos^2(\alpha)) \Longrightarrow (50)/(49) =(1)/(\cos^2(\alpha)) $`

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

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

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

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)`

`$\cos(2\beta)=(1-\tan^2(\beta))/(1+\tan^2(\beta)) =(1-((1)/(7))^2 )/(1+((1)/(7))^2)=(24)/(25)$`

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

`$\sin(2\beta)=2\sin(\beta)\cos(\beta)=2 \cdot (√(10) )/(10)\cdot (3√(10) )/(10) = (3)/(5) $`

Now we can perform the sum identity:

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

`$\sin(\alpha + 2\beta)=(√(2))/(10)\cdot (24)/(25) +(3)/(5) \cdot (7√(2) )/(10) = (129√(2))/(250)$`

But we are not done yet! You want

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

You actually want the

`$\arcsin\left((129√(2))/(250)\right)\approx 0.8179$`