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Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?

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  • $\begingroup$ This term is equivalent to $6\cot^{2} x + 4 \tan^{2} x$ , hoping no quick error. Done it in a hurry. Where min is 10 $\endgroup$ – Mann May 13 '15 at 17:35
  • $\begingroup$ Could you please tell me how did you simplify? $\endgroup$ – user167045 May 13 '15 at 17:39
  • $\begingroup$ I differentiated to get the answer as 9.798 $\endgroup$ – Hiten May 13 '15 at 17:44
  • $\begingroup$ The middle term doesn't matter. Since $\sin^2 2x=4\sin^2 x\cos^2 x$ we need to minimize $(4+2t^2)/(t(1-t)$ where $t=\cos^2 x$. Routine calculation.. $\endgroup$ – André Nicolas May 13 '15 at 17:57
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$$8\cos^4x = 8\left(\dfrac{1+\cos (2x)}{2}\right)^2 = 8\left(\dfrac{1+2\cos (2x)+ \cos^2(2x)}{4}\right)=2+4\cos (2x)+2\cos^2(2x) = 2+4t+2t^2, t = \cos (2x) \Rightarrow y = \dfrac{16-8(1-t^2)+2+4t+2t^2}{1-t^2} = \dfrac{10+10t^2+4t}{1-t^2}=f(t), -1 \leq t \leq 1$$. Can you take it from here?

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$$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x}$$

or $$y=\frac{16}{4}*\sec^2 x \csc^2 x-8 +\frac{8}{4}\frac{\cos^{4} x}{\sin^2 x* \cos^2 x}$$

or

$$y=4(\tan^2 x +1)(\cot^2 x +1)-8+2 \cot^2 x$$

$$y=4(\tan^2 x + \cot^2 x + 2)-8+2 \cot^2 x$$

Which after simplifying gives,

$$y=4\tan^2 x + 6\cot^2 x$$

Edit: as user suggested the answer in comment is not valid. But one here can easily use AM-GM inequality to reach at correct answer

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  • $\begingroup$ Minimum of course at $x=\pi/4$ $\endgroup$ – Mann May 13 '15 at 17:44
  • $\begingroup$ @ Mann : No, it is not so. See my answer done with Maple. $\endgroup$ – user64494 May 13 '15 at 18:59
  • $\begingroup$ Or well maybe close to it xD $\endgroup$ – Mann May 13 '15 at 19:11

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