Show that

$$\dfrac{3+\cos x}{\sin x}\quad \forall \quad x\in R $$

cannot have any value between $-2\sqrt{2}$ and $2\sqrt{2}$.

My attempt is as follows:

There can be four cases, either $x$ lies in the first quadrant, second, third or fourth:-

First quadrant: $\cos x$ will decrease sharply and sinx will increase sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.

$y_{max}$ would tend to $\infty$ near to $x=0$

Second quadrant: $\cos x$ will increase in magnitude and sinx will decrease sharply, so $y_{min}=3$ at $x=\dfrac{\pi}{2}$.

$y_{max}$ would tend to $\infty$ near to $x=\pi$

Third quadrant: $\cos x$ will decrease in magnitude and sinx will increase in magnitude but negative, so $y_{min}$ would tend to $-\infty$ near to $x=\pi$

$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$

Fourth quadrant: $\cos x$ will increase sharply and sinx will decrease in magnitude, so $y_{min}$ would tend to $-\infty$ near to $x=2\pi$

$y_{max}$ would be $-3$ at $x=\dfrac{3\pi}{2}$

So in this way I have proved that $\dfrac{3+\cos x}{\sin x}$ cannot lie between $-2\sqrt{2}$ and $2\sqrt{2}$, but is their any smart solution so that we can calculate quickly.

  • 1
    $\begingroup$ It may help to rewrite the expression as $$3\csc x + \cot x. $$ $\endgroup$ Oct 28, 2019 at 1:02
  • $\begingroup$ yeah I tried that but how will you calculate the range after that $\endgroup$ Oct 28, 2019 at 1:03
  • 1
    $\begingroup$ First and second derivative tests for $x\in[0,2\pi]$, I suppose. $\endgroup$ Oct 28, 2019 at 1:04
  • 1
    $\begingroup$ Substitute $\sin(x)=\frac{2t}{1+t^2}$ and $\cos(x)=\frac{1-t^2}{1+t^2}$. The expression becomes $y=\frac{2+t^2}{t}$. You can find for what values of $y$ there is a corresponding $t$ by looking at the discriminant of $t^2-yt+2=0$, which is $y^2-8$. So, for $|y|<2\sqrt{2}$ there are no solutions. $\endgroup$ Oct 28, 2019 at 1:25

4 Answers 4


Use the half-angle expressions $\cos x = \frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}$ and $\sin x = \frac{2\tan\frac x2}{1+\tan^2\frac x2}$ to express

$$I=\frac{3+\cos x}{\sin x}= \frac{2}{\tan\frac x2} +\tan \frac x2$$


$$I^2=\left(\frac{2}{\tan \frac x2} +\tan \frac x2\right)^2 =\left(\frac{2}{\tan \frac x2} -\tan \frac x2\right)^2+8 \ge 8$$

Thus, $I^2$ can not have values within $[0,8)$, which means that $I=\frac{3+\cos x}{\sin x}$ can not have values within $(-2\sqrt2, \>2\sqrt2)$.


Let the value of this expression at $x=p$ be $q$. Then we have

$${3+\cos p\over\sin p}= q$$ $$\implies\cos p-q\sin p = -3$$ Dividing both LHS and RHS by $\sqrt{q^2+1}$ we have

$$\cos p\cdot {1\over\sqrt{q^2+1}} - \sin p\cdot{q\over\sqrt{q^2+1}} = {-3\over\sqrt{q^2+1}}$$

Let $1\over\sqrt{q^2+1}$ be $\cos r$. So, we have $$\cos p\cos r - \sin p\sin r = {-3\over\sqrt{q^2+1}}$$

Or $$\cos{(p+r)} = \frac{-3}{\sqrt{q^2+1}}$$

For this to be a valid expression $\sqrt{q^2+1}$ must be greater than $3$ since $\cos x \in [-1,1]$. So, we have

$$q^2+1 \geq 9$$ $$\implies |q| \geq \sqrt 8$$ $$\implies q \in \left(-\infty,-2\sqrt 2\right]\cup\left[2\sqrt 2, \infty\right)$$ $$\implies\boxed{ {3+\cos x\over\sin x}\notin\left(-2\sqrt 2, 2\sqrt 2\right)}$$


Basically you want to prove: $$\left|\dfrac{3+\cos x}{\sin x}\right|\ge 2\sqrt{2} \iff (3+\cos x)^2\ge 8\sin ^2x \iff (3\cos x+1)^2\ge 0 \ \ \checkmark$$ Note: Equality occurs for $\cos x=-\frac13 \Rightarrow \sin x=\pm \frac{2\sqrt2}{3}$.


If $\frac{3 + \cos x}{\sin x} = k, k \in \mathbb R$, then $3 + \cos x = k \sin x \Rightarrow 9 + 6 \cos x + \cos^2 x = k^2 - k^2 \cos^2 x$, thus $(k^2+1)\cos^2 x + 6 \cos x + (9 - k^2) = 0$. Let $u = k^2$. Now for no value of $\cos x$ to exist, the discriminant must be less than $0$:

$$6^2 - 4(u+1)(9-u) < 0 \Rightarrow 4(u+1)(9-u) > 36$$ $$\Rightarrow -u^2 + 8u + 9 > 9 \Rightarrow u(-u+8) > 0 \Rightarrow 0 < u < 8$$

where in the last step, a sketch of the quadratic shows that it is concave up, hence the direction of the inequality.

Thus $0 < k^2 < 8$. This implies there are no values of $x$ such that $-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$.


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