Show that $\frac{3\> + \>\cos x}{\sin x}$ cannot have any value between $-2\sqrt2$ and $2\sqrt2$

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.

• It may help to rewrite the expression as $$3\csc x + \cot x.$$ Oct 28, 2019 at 1:02
• yeah I tried that but how will you calculate the range after that Oct 28, 2019 at 1:03
• First and second derivative tests for $x\in[0,2\pi]$, I suppose. Oct 28, 2019 at 1:04
• 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. Oct 28, 2019 at 1:25

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

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

Note

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

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 $$-2 \sqrt{2} < \frac{3 + \cos x}{\sin x} < 2 \sqrt{2}$$.