Show that $$y=\dfrac{4\sin\theta}{2+\cos\theta}-\theta$$ is increasing function when $\theta \in [0,\frac\pi2]$

What I have done

If $\theta_1,\theta_2\in[0,\frac\pi2]$

then $$\sin\theta_1 < \sin\theta_2 \implies 4\sin\theta_1 < 4\sin\theta_2$$ $$\cos\theta_1 > \cos\theta_2 \implies 2+\cos\theta_1 > 2+\cos\theta_2$$

In $\frac{4\sin\theta}{2+\cos\theta}$ we see that as $\theta$ increases $4\sin\theta$ increases and $2+\cos\theta$ decreases.

$\therefore$ $\frac{4\sin\theta}{2+\cos\theta}$ increases as $\theta$ increases.

What problem I have

I think that if I show that $$\frac{d}{d\theta}(\frac{4\sin\theta}{2+\cos\theta}) > \frac{d\theta}{d\theta}=1$$

then the solution will be completed.

Is my technique correct?
If yes then please help me in showing $$\dfrac{d}{d\theta}\left(\frac{4\sin\theta}{2+\cos\theta}\right)>1.$$ If no then kindly tell me where I'm wrong.

  • $\begingroup$ You can differentiate using the Quotient Rule and obtain an expression for the derivative of the function which is strictly positive on the interval you are considering $\endgroup$ – David Quinn Jul 27 '15 at 9:08
  • $\begingroup$ Sir I found the derivative and it is strictly positive(more than zero) but I have to show that, that derivative is more than 1. Sir please help me to show that derivative is more than 1. $\endgroup$ – Singh Jul 27 '15 at 9:14
  • $\begingroup$ See the answer below from Vinod $\endgroup$ – David Quinn Jul 27 '15 at 9:26
  • $\begingroup$ $$(y+\theta)(2+\cos\theta)=4\sin\theta$$ You can differentiate this using product formula. $\endgroup$ – Bumblebee Jul 27 '15 at 9:37


$4-\cos\theta>0$ because $\cos\theta$ is always less than $1$, therefore the function is increasing in $\left[0,\frac{\pi}{2}\right]$

  • 1
    $\begingroup$ Thank you sir. I just forgot to put $4\cos^2\theta+4\sin^2\theta=4$. Lack of experience and focus! $\endgroup$ – Singh Jul 27 '15 at 9:45

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