How to prove (analytically) that $\sin(\tan^{-1}(\sqrt{2}))=\sqrt{\frac{2}{3}}$?

Question

I calculated $\sin(\tan^{-1}(\sqrt{2}))$ using Wolfram and got it as a result $\sqrt{\frac{2}{3}}$. However, I was unable to find an identity to obtain this result.

Is there any way to calculate this value without using Wolfram?

Answer 1

Let $t=\tan^{-1} \sqrt2$. Then, $ \tan t=\sqrt2$ and $$\sin t = \tan t\cos t= \frac {\tan t}{\sqrt{1+\tan^2t}}=\sqrt{\frac23} $$

Answer 2

Well, $\arctan x$ is an angle whose tangent function is $\frac{x}{1}$. Considering the sides of the right triangle. We have apposite side is $x$, adjacent side is $1$ and hypothenuse is $\sqrt{1+x^2}$. Therefore the sine of this angle is:

$$\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{x}{\sqrt{1+x^2}}\tag1$$

So, we have:

$$\sin\left(\arctan\left(x\right)\right)=\frac{x}{\sqrt{1+x^2}}\tag2$$

In your case:

$$\sin\left(\arctan\left(\sqrt{2}\right)\right)=\frac{\sqrt{2}}{\sqrt{1+\left(\sqrt{2}\right)^2}}=\frac{\sqrt{2}}{\sqrt{3}}=\sqrt{\frac{2}{3}}\tag3$$

Answer 3

Hint:

Use the mid-school formula: $$\sin^2\theta=\frac{\tan^2\theta}{1+\tan^2\theta}$$ and use that $\sin\theta$ and $\tan\theta$ have the same sign on $\bigl(-\frac\pi 2,\frac\pi 2\bigr)$.

Answer 4

Also, the angle $t$ of right triangle satisfies

$$\sin^2 t= \dfrac{2}{3}$$ $$\cos^2 t= \dfrac{1}{3}$$ $$\tan^2 t= 2 $$