How do you evaluate $\sin \left( 2\arctan \left( \sqrt{2} \right) \right)?$

Hint: We use trigonometric and inverse trigonometric identities to evaluate $\sin \left( 2\arctan \left( \sqrt{2} \right) \right).$ Let us recall the identity: $\sin \left( \arctan \left( x \right) \right)=\dfrac{x}{\sqrt{1+{{x}^{2}}}}.$ Also remember the identity: $\cos \left( \arctan \left( x \right) \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}.$ The basic trigonometric identity $\tan x=\dfrac{\sin x}{\cos x}$ will be useful while evaluating the given trigonometric function. We may use the trigonometric identity, $\sin 2x=2\sin x\cos x.$

Complete step by step solution:
Consider the given trigonometric function $\sin \left( 2\arctan \left( \sqrt{2} \right) \right).$
We are going to start with the function $\arctan \left( \sqrt{2} \right).$
Suppose that $x=\arctan \left( \sqrt{2} \right).$
On the basis of our supposition, we deduce that $\tan x=\sqrt{2}.$
Therefore, what we have to find is the value or values of $x$ for which $\tan x=\sqrt{2}$ is true.
We know the trigonometric identity that connects sine, cosine and tangent as $\tan x=\dfrac{\sin x}{\cos x}.$
From this we will get the following,
$\Rightarrow \dfrac{\sin x}{\cos x}=\sqrt{2}.$
That is, we have to find the value or values of $x$ for which the quotient $\dfrac{\sin x}{\cos x}$ is equal to $\sqrt{2}.$
Also, we should keep the fact that the sign of both the trigonometric functions sine and cosine should be the same. Because, the value of the tangent function is positive.
Now, we can say that $\tan x=\sqrt{2}$ implies,
 either $\sin x=\dfrac{\sqrt{2}}{\sqrt{3}}$ and $\cos x=\dfrac{1}{\sqrt{3}}$
or $\sin x=-\dfrac{\sqrt{2}}{\sqrt{3}}$ and $\cos x=-\dfrac{1}{\sqrt{3}}.$
Now let us go back to the given trigonometric function.
So, we will have $\sin \left( 2\arctan \left( \sqrt{2} \right) \right)=\sin \left( 2x \right),$ we have, earlier, mentioned that $x=\arctan \left( \sqrt{2} \right).$
Now we use another trigonometric identity to evaluate the above obtained equation.
That is, $\sin 2x=2\sin x\cos x.$
We are going to apply this identity to our problem to get,
\[\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=\sin 2x=2\sin x\cos x.\]
Let us substitute the values of the sine function and the cosine function.
We get,
\[\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=2\dfrac{\sqrt{2}}{\sqrt{3}}\dfrac{1}{\sqrt{3}}.\]
Or,
\[\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=2\dfrac{-\sqrt{2}}{\sqrt{3}}\dfrac{-1}{\sqrt{3}}.\]
In both the cases, we get
\[\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=2\dfrac{\sqrt{2}.1}{\sqrt{3}.\sqrt{3}}.\]
Hence, we get,
\[\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=2\dfrac{\sqrt{2}}{3}.\]
Hence, the simplified form of \[\sin \left( 2\arctan \left( \sqrt{2} \right) \right)=2\dfrac{\sqrt{2}}{3}.\]

Note: If we know $\tan x=\sqrt{2}$ when,
$\sin x=\dfrac{\sqrt{2}}{\sqrt{3}}$ and $\cos x=\dfrac{1}{\sqrt{3}}$
or $\sin x=-\dfrac{\sqrt{2}}{\sqrt{3}}$ and $\cos x=-\dfrac{1}{\sqrt{3}},$
then, we can directly find the value of the given function using the identity given below:
$2\arctan x=\arcsin \left( \dfrac{2x}{1+{{x}^{2}}} \right),\,\,\,\left| x \right|\le 1.$
Take $x=\sqrt{2},$ and then we will get,
$\Rightarrow 2\arctan \sqrt{2}=\arcsin \left( \dfrac{2\sqrt{2}}{1+{{\left( \sqrt{2} \right)}^{2}}} \right),\,\,\,\left| \sqrt{2} \right|\le 1.$
This will become,
$\Rightarrow 2\arctan \sqrt{2}=\arcsin \left( \dfrac{2\sqrt{2}}{1+2} \right),\,\,\,\left| \sqrt{2} \right|\le 1.$
Thus, we get,
$\Rightarrow 2\arctan \sqrt{2}=\arcsin \left( \dfrac{2\sqrt{2}}{3} \right),\,\,\,\left| \sqrt{2} \right|\le 1.$
Hence, we get
$\Rightarrow \sin \left( 2\arctan \left( \sqrt{2} \right) \right)=\left( \dfrac{2\sqrt{2}}{3} \right),\,\,\,\left| \sqrt{2} \right|\le 1.$