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Equation: $\sin(2\arctan x)$

And I have to simplify it. I can only go this far, and then I lack ideas:

Let $\arctan x = \alpha$

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

So I haven't progressed really.

Perhaps I should use $\cos^2 \alpha + \sin^2 \alpha= 1$, to express $\tan$ as $\mathrm{cosec}$ and then as $\sec$ and insert those into the above. But I have no clue what $\sin(\mathrm{acosec}(x))$ would equal to (or $\cos(\mathrm{arcsec}(x))$).

EDIT: I see now how to express $\sin$ or $\cos$ of an arbitrary angle just in terms of tangent.

$$\tan \mu \equiv \frac{\sin \mu}{\cos \mu}$$ $$\sin \mu \equiv \tan \mu \cos \mu$$ $$\cos \mu \equiv \frac{1}{\sec \mu}$$ $$\tan^2 \mu + 1 \equiv \sec^2 \mu$$ $$\sec \mu \equiv \sqrt{\tan^2 \mu + 1}$$ Therefore, $$\sin \mu \equiv \frac{\tan \mu}{\sqrt{\tan^2 \mu + 1}}$$

Also, there is a nice way of expressing $\sin 2\mu$ in terms of $\tan \mu$ which is extremely useful in this case. It can be found in the answer that I have selected.

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3 Answers 3

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Hint:

$\arctan x=\alpha$ means that $ x=\tan \alpha$. So you starting point is correct, simply you have to use the classical formulas:

$$ \sin \alpha= \pm \dfrac{\tan \alpha}{\sqrt{1+\tan^2 \alpha}} $$

$$ \cos \alpha= \pm \dfrac{1}{\sqrt{1+\tan^2 \alpha}} $$ that becomes: $$ \sin \alpha= \pm \dfrac{x}{\sqrt{1+x^2}} $$

$$ \cos \alpha= \pm \dfrac{1}{\sqrt{1+x^2}} $$

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If $\arctan x=u\implies x=\tan u$

$$\sin(2\arctan x)=\sin2u=\dfrac{2\tan u}{1+\tan^2u}=?$$

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Let $u=\arctan x$ ; we have $$\sin\left(\arctan x\right)=\sin u=\sqrt{1-\cos^2 u}=\sqrt{1-\frac{1}{1+\tan^2 u}}=\frac{\tan u}{1+\tan^2 u}=\frac{x}{\sqrt{1+x^2}}.$$ Similarly, we have $$\cos\left(\arctan x\right)=\frac{1}{\sqrt{1+x^2}}.$$ Thus, $$\sin\left(2\arctan x\right)=2\sin\left(\arctan x\right)\cos\left(\arctan x\right)=\frac{2x}{1+x^2}.$$

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