For $0<\theta<\pi/6$ all the values of the expression

$\tan^23\theta \cos^2\theta-4\tan3\theta \sin2\theta+16\sin^2\theta$ lies in what interval.

I actually took $\sin^2\theta$ common out of the expression to yield,

$$(\frac{\tan^23\theta}{\tan^2\theta}-\frac{8\tan3\theta}{\tan\theta} +16)\sin^2\theta$$

$$=(\frac{\tan3\theta}{\tan\theta} -4)^2\sin^2\theta$$

Now how can I now proceed further ahead.

  • $\begingroup$ Personally, I think you should just find sup and inf of that function on $\left(0,\frac\pi6\right)$ and see if they are is attained for some $\theta$. $\endgroup$ – user228113 Jun 13 '16 at 18:29

$$(\frac{tan3\theta}{tan\theta} -4)^2sin^2\theta$$ $$=(\frac{11tan^2\theta-1}{1-3tan^2\theta} )^2sin^2\theta$$

Now $sin$ is an increasing function in the domain.

Also if you check the derivative(s) (or rough graphs of$\frac{11tan^2\theta-1}{1-3tan^2\theta}$) you will notice that $\frac{11tan^2\theta-1}{1-3tan^2\theta}$ is symmetric about $0$ (and value at $\theta=0$ is $-1$ ) and has a root at (say) $0\lt x_0\lt \pi/6$ (at $\theta=\pi/6 $ its $\infty$). So squaring the thing will just reflect the part below $X$ axis above , keeping the root intact . So minimum value of the expression is zero (achieved at 2 points $0\ \&\ x_0$) and maximum is infinity

  • $\begingroup$ @HarshSharma As a member of this site, I request you not to leave answers unaccepted .(A bad habit that has been incurred by bout 50% people) . If you dont want to accept an answer, please give a reason.. $\endgroup$ – Qwerty Jun 15 '16 at 4:28

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