Need help simplifying a trigonometric expression

**nubshat** · October 6th 2012, 09:34 AM

Cot(2arcsinx)

Thanks in advance

**Siron** · October 6th 2012, 10:22 AM

Note that $\cot(2\arcsin x) = \frac{1}{\tan(2\arcsin x)}$

From this point you can use the formula $\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}$ and $\tan(\arcsin x) = \frac{\sin(\arcsin x)}{\cos(\arcsin x)} = \frac{x}{\sqrt{1-x^2}}$

**Soroban** · October 6th 2012, 11:52 AM

Hello, nubshat!

$\text{Simplify: }\cot(2\arcsin x)$

Let $\theta \,=\,\arcsin x\quad\Rightarrow\quad \sin\theta \,=\,x$

That is: . $\sin\theta \:=\:\frac{x}{1}\:=\:\frac{opp}{hyp}$

$\theta$ is in a right triangle with: $opp = x$ and $hyp = 1.$

Pythagorus says: . $adj \,=\,\sqrt{1-x^2}$

Hence: . $\cot\thet \,=\,\frac{adj}{opp} \,=\,\frac{\sqrt{1-x^2}}{x}$

Identity: . $\cos2\theta \:=\:\frac{\cot^2\theta - 1}{2\cot\theta}$

We have: . $\cot2\theta \:=\:\frac{\left(\frac{\sqrt{1-x^2}}{x}\right)^2-1}{2\left(\frac{\sqrt{1-x^2}}{x}\right)} \;=\; \frac{\frac{1-x^2}{x^2} - 1}{\frac{2\sqrt{1-x^2}}{x}}$

Multiply by $\tfrac{x^2}{x^2}\!:\;\;\frac{x^2}{x^2}\cdot\frac{ \frac{1-x^2}{x^2} - 1}{\frac{2\sqrt{1-x^2}}{x}} \;=\;\frac{1-x^2-x^2}{2x\sqrt{1-x^2}}$

Therefore: . $\cos2\theta \;=\;\frac{1-2x^2}{2x\sqrt{1-x^2}}$

**StefanTM** · October 8th 2012, 07:43 AM

Hi Soroban,

a little attention error in the last row:

cot (2arcsin(x)) = ....
and not
cos(2arcsin(x))=...

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