Finding the exact value of an inverse trigonometric expression?

[crayolakat]

Here is the problem I'm stuck on:

sin(2tan^-1sqrt(2))

To find the exact value of this function, I was pretty sure that I had to change the tan^-1sqrt(2) to tanx=sqrt(2).

However, that seems impossible. I cannot find any value of x that would make tanx=sqrt(2).

Am I approaching this problem wrong, or am I missing something here?

 

[Deleted member 4993]

Here is the problem I'm stuck on:

sin(2tan^-1sqrt(2))

To find the exact value of this function, I was pretty sure that I had to change the tan^-1sqrt(2) to tanx=sqrt(2).

However, that seems impossible. I cannot find any value of x that would make tanx=sqrt(2).

Am I approaching this problem wrong, or am I missing something here?

tanΘ=sqrt(2) → sinΘ = √(2/3) and cosΘ = √(1/3) assuming 0 ≤ 2Θ ≤ π/2 ..........................................................(1)


sin(2tan^-1sqrt(2)) = sin(2Θ) = 2 * sinΘ * cosΘ .........................and continue

 

[MarkFL]

You want to use the double-angle identity for sine to write:

\(\displaystyle \sin\left(2\tan^{-1}(\sqrt{2}) \right)=2\sin\left(\tan^{-1}(\sqrt{2}) \right)\cos\left(\tan^{-1}(\sqrt{2}) \right)\)

Now, you want to consider an angle in a right triangle that has \(\displaystyle \sqrt{2}\) as the opposite side and 1 as the adjacent side. What would the hypotenuse be?

 

[crayolakat]

tanΘ=sqrt(2) → sinΘ = √(2/3) and cosΘ = √(1/3) assuming 0 ≤ 2Θ ≤ π/2 ..........................................................(1)


sin(2tan^-1sqrt(2)) = sin(2Θ) = 2 * sinΘ * cosΘ .........................and continue

So, 2*((sqrt2)/3)(1/3), correct?

That would give me (2/9)(sqrt2).

The answer is supposed to be (2/3)(sqrt2), what am I still doing something wrong

Never mind. I understand, thank you