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I faced this formula in Analytic Geometry course and looking for ways to derive it. All I know is this $\sin(2x) = 2\sin(x) \cos(x) $Can you help me.

$$ \sin(2x) =\frac{1}{ \sqrt{1 + \cot^2(2x)} }$$

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    You can't derive the formula, because it is false. Take $x=2\pi/3$ to see why: the left-hand side is negative, whereas the right-hand side is positive. You need $\lvert\sin(2x)\rvert$ in the left-hand side. – 2017-01-08
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    You must take the absolute value of the LHS, and stipulate that $\sin 2\theta \ne 0$ for this to be true. – 2017-01-08

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Hint:

$$1+\cot^2x=1+\frac{\cos^2x}{\sin^2x}=\frac{\sin^2x+\cos^2x}{\sin^2x}=\frac1{\sin^2x}$$

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    I got the point. But how did you know that $1+\cot^2x$ will result in explanation? – 2017-01-08
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    @BaimyrzaShamyr What? What do you mean by that? I don't understand... – 2017-01-08
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I start with the expression $\frac {1}{\sin 2x} $. We thus have $$E = \frac {1}{\sin 2x} \Rightarrow E^2 =\frac{1}{\sin^2 2x}= \frac {\sin^2 2x +\cos^2 2x}{\sin^2 2x}= 1+ \cot^2 2x $$ Thus, $$\frac {1}{E} =\sin 2x =\frac {1}{\sqrt {1+\cot^2 2x}} $$ But we can see that the LHS can take negative values but the RHS cannot. So we extend the domain of the LHS by putting the absolute value to the LHS.

Thus, more rightly, $$\frac {1}{E} =|\sin 2x| =\frac {1}{\sqrt {1+\cot^2 2x}} $$ Hope it helps.

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    Do you think that $\sqrt{E^2}=E$? – 2017-01-08
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It is better to solve RHS as it is more complex. Notice that $1+\cot^2(x)=\csc^2(x)$ so $\frac{1}{ \sqrt{1 + \cot^2(2x)} }=\frac{1}{\sqrt{\csc^2(2x)}}=|\sin (2x)|$

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$$|{\sin 2x}|= \frac{1}{\csc 2x} = \frac{1}{\sqrt{ (\csc 2x)^2}} = \frac{1}{\sqrt{1+\cot^2 2x}}$$

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    Still wrong: not the secant, but the cosecant. However, the major mistake $x=\sqrt{x^2}$ is still there. – 2017-01-08
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    @egreg thanks for noting my mistake,i think we can write $x=\sqrt{x^2}$ till x is positive number – 2017-01-08
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    Yes, so you can derive the correct formula, namely $\lvert\sin2x\rvert=\dfrac{1}{\sqrt{1+\cot^22x}}$ – 2017-01-08
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1.$\frac{1}{\sqrt{1+cot^2(2x)}}=\frac{1}{\sqrt{1+\frac{cos^2(2x)}{sin^2(2x)}}}$

2.$\frac{1}{\sqrt{1+\frac{cos^2(2x)}{sin^2(2x)}}}=\frac{1}{\sqrt{\frac{sin^2(2x)+cos^2(2x)}{sin^2(2x)}}}$

3.$\frac{1}{\sqrt{\frac{sin^2(2x)+cos^2(2x)}{sin^2(2x)}}}=\frac{1}{\sqrt{\frac{1}{sin^2(2x)}}}$*

4.$\frac{1}{\sqrt{\frac{1}{sin^2(2x)}}}=\frac{1}{\frac{1}{sin(2x)}}=sin(2x)$

*(here I used the identity $sin^2(2x)+cos^2(2x)=1$)

**(I have taken the positive square root in fourth step.)

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You can do something like this: $$\sin(2x)=\sqrt{sin^2(2x)}$$ $$=\sqrt{\frac{1}{cosec^2(2x)}}$$ $$=\sqrt{\frac{1}{1+cot^2(2x)}}\tag{$\because cosec^2\theta=1+cot^2\theta$}$$