Question 1120469
<font face="Times New Roman" size="+2">


Not complicated at all, actually.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin^2(2x)\[\cot^2(x)\ -\ \tan^2(x)\]]


Double angle formula for sine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(2sin(x)cos(x)\)^2\[\cot^2(x)\ -\ \tan^2(x)\]]


tangent is sine over cosine and cotangent is the reciprocal of tangent:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(2sin(x)cos(x)\)^2\[\frac{\cos^2(x)}{\sin^2(x)}\ -\ \frac{sin^2(x)}{\cos^2(x)}\]]


Apply common denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(2sin(x)cos(x)\)^2\[\frac{\cos^2(x)\ -\ \sin^2(x)}{\sin^2(x)\cos^2(x)}\]]


Square the first factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(4sin^2(x)cos^2(x)\)\[\frac{\cos^2(x)\ -\ \sin^2(x)}{\sin^2(x)\cos^2(x)}\]]


Eliminate factors common to numerator and denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\(\cos^2(x)\ -\ \sin^2(x)\)]


Apply double angle identity for cosine:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\cos(2x)]


Q.E.D.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>