# Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$

Hi this is the question:

Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$

What I've tried:

I know that the range of $$\csc(x)$$ which is $$R\setminus (-1,1)$$, the range of $$\sec(x)$$ is $$R\setminus (-1,1)$$ too. And I've managed to simplify the expression to have in terms of $$\sec x$$ and $$\csc x$$ as such: $$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$ But then, what do I do when I have 3 terms and not just $$\sec$$ or $$\csc$$?

$$f(x)=\dfrac{1+\sin x+\cos x}{\sin x\cos x}$$

Clearly $$f(x)$$ is undefined for $$2x=n\pi$$ where $$n$$ is any integer

$$\dfrac{f(x)}2=\dfrac{1+\sin x+\cos x}{(\sin x+\cos x)^2-1}=\dfrac1{\sin x+\cos x-1}$$

Now use $$-\sqrt2\le\sin x+\cos x\le\sqrt2$$ to find

$$\implies-\sqrt2-1\le\sin x+\cos x-1\le\sqrt2-1$$

If $$\sin x+\cos x-1\ge0, f(x)\ge2+2\sqrt2$$

and if $$\sin x+\cos x-1\le0,f(x)\le2-2\sqrt2$$

• What is that? Is there any other method Commented Oct 29, 2020 at 10:32
• @Xetrez, Please pinpoint your confusion ? Commented Oct 29, 2020 at 10:35
• How did you go from 1st to 2nd Commented Oct 29, 2020 at 10:44
• @Xetrez, $$(\sin x+\cos x)^2-1=?$$ Commented Oct 29, 2020 at 10:45
• $1+2\sin(x)\cos(x)-1=2\sin(x)\cos(x)$ Commented Oct 29, 2020 at 10:48

The derivative of the secant function is $$\sin x/\cos^2x$$; the derivative of the cosecant function is $$-\cos x/\sin^2x$$, so you get $$f'(x)=-4\frac{\cos2x}{\sin^22x}+\frac{\sin x}{\cos^2x}-\frac{\cos x}{\sin^2x} =\frac{\sin^2x-\cos^2x+\sin^3x-\cos^3x}{\sin^2x\cos^2x}$$ We can disregard the denominator and factor the numerator as $$(\sin x-\cos x)(\sin x+\cos x+\sin^2x+\sin x\cos x+\cos^2x) =(\sin x-\cos x)(1+\sin x)(1+\cos x)$$ Quite nicer! This is positive where $$\sin x>\cos x$$. Limiting ourselves to the interval $$(0,2\pi)$$, but also noting that the function is undefined at integer multiples of $$\pi2$$, we see that this happens for $$\pi/4. Thus our function is

• decreasing over $$(0,\pi/4)$$
• increasing over $$(\pi/4,\pi/2)$$
• increasing over $$(\pi/2,5\pi/4)$$
• decreasing over $$(5\pi/4,3\pi/2)$$
• decreasing over $$(3\pi/2,2\pi)$$

Thus, taking into account the asymptotes, we see that the branch in $$(0,\pi/2)$$ has range $$(f(\pi/4),\infty)=(2+2\sqrt{2},\infty)$$.

The branch in $$(\pi/2,2\pi)$$ is a bit more complicated, because the function doesn't have asymptotes at $$\pi$$ and at $$3\pi/2$$. Indeed, $$\lim_{x\to\pi}f(x)=-1=\lim_{x\to3\pi/2}f(x)$$ If we extend the function by continuity, the range over $$(\pi/2,2\pi)$$ would be $$(-\infty,f(5\pi/4))=(-\infty,2-2\sqrt{2})$$.

As the function is not defined at $$\pi$$ and $$3\pi/2$$, the range is $$(-\infty,-1)\cup(-1,2-2\sqrt{2})\cup(2+2\sqrt{2},\infty)$$

• Why do we disregard the denominator? Commented Oct 29, 2020 at 19:07

Starting with $$f(x)=\frac{2}{\sin x+ \cos x-1} \implies y=\frac{1+t^2}{t-t^2} \implies (1+y)t^2-yt-1=0,$$ As $$t=\tan(x/2)$$ lies in $$(-\infty, \infty)$$, the range will be all values of $$y$$ when the above quadratic has real roots: $$B^2 \ge 4AC$$ $$\implies y^2-4y-4 \ge 0 \implies y\ge 2+\sqrt{2} ~or~ y\le2-\sqrt{2}$$

First of all, we need

$$\csc2x$$ must be finite $$\implies\sin2x\ne0\implies2x\ne n\pi\ \ \ \ (1)$$ where $$n$$ is any integer

Now if $$\sin x+\cos x+1=0,$$

using double angle formula, $$\cos\dfrac x2\left(\sin\dfrac x2+\cos\dfrac x2\right)=0$$

$$\cos\dfrac x2=0\iff x=(2r+1)\pi\ \ \ \ (2)$$

and $$\sin\dfrac x2+\cos\dfrac x2=0\implies x=2r\pi-\dfrac\pi2\ \ \ \ (3)$$

Both $$(2),(3)$$ are been precluded by $$(1)$$

Let $$\sin x+\cos x+1=k\implies k\ne0$$

$$\sin x\cos x=\dfrac{(\sin x+\cos x)^2-1}2=\dfrac{k^2-2k}2$$

$$f(x)=\dfrac{2k}{k(k-2)}=\dfrac2{k-2}$$

As $$k\ne0, f(x)\ne\dfrac2{0-2}\ \ \ \ (4)$$

Now $$-\sqrt2\le\sin x+\cos x\le\sqrt2\implies-\sqrt2-1\le k-2\le\sqrt2-1$$

If $$0< k\le\sqrt2-1, f(x)\ge\dfrac2{\sqrt2-1}=?$$

If $$0>k\ge-\sqrt2-1, f(x)\le\dfrac2{-(\sqrt2+1)}=?$$

But remember $$(4)$$

• I don't understand this at all, where's the other solution you sent? Commented Oct 29, 2020 at 17:25
• @Xetrez, I have deleted that, tried to put a cleaner solution! Where is your confusion Commented Oct 29, 2020 at 18:01
• Why did you do so? It was so much simoler than THAT. I was only confused by some steps, what was wrong with the other one though? And this as far as i can comprehend, this looks way anove my level Commented Oct 29, 2020 at 19:02
• @Xetrez, Undeleted temporarily. Please grasp basic trigonometric identities Commented Oct 29, 2020 at 20:05
• alright thanks so much Commented Oct 29, 2020 at 20:49