# TrigonometryRange of sec^4(x) + csc^4(x)

#### abender

##### New member
Hi, fellas. In response to someone's request on another form to find the range of $$\sec^4(x)+\csc^4(x)$$, I offered the following solution and explanation:

[TEX]\sec^4(x) = (\sec^2(x))^2 = (1+\tan^2(x))^2 = \tan^4(x) + 2\tan^2 + 1[/TEX]

[TEX]\csc^4(x) = (\csc^2(x))^2 = (1+\cot^2(x))^2 = \cot^4(x) + 2\cot^2 + 1 [/TEX]

[TEX]\sec^4(x) + \csc^4(x) = \tan^4(x) + 2\tan^2 + 1 + \cot^4(x) + 2\cot^2 + 1 [/TEX]$$= \tan^4(x) + \cot^4(x) + 2(\tan^2(x)+\cot^2(x)) + 2$$

The range of [TEX]\tan(x)[/TEX] is [TEX](-\infty, \infty)[/TEX]. Likewise, the range of [TEX]\cot(x)[/TEX] is [TEX](-\infty, \infty)[/TEX].

Since [TEX]\tan^4(x)[/TEX] and [TEX]\cot^4(x)[/TEX] are positive even powers, both have range [TEX][0,\infty)[/TEX].

BELOW is where others may disagree with me:

I contend that the range of [TEX]\tan^4(x)+\cot^4(x)[/TEX] is [TEX](0,\infty)[/TEX] as opposed to [TEX][0,\infty)[/TEX], which the sum of the parts may intuitively suggest.
The two ranges differ insofar [TEX](0,\infty)[/TEX] does not contain [TEX]0[/TEX], whereas [TEX][0,\infty)[/TEX] does contain [TEX]0[/TEX].
I believe that the range of [TEX]\tan^4(x)+\cot^4(x)[/TEX] should NOT include 0, i.e., it should be [TEX](0,\infty)[/TEX].

Proof is achieved if we show [TEX]\tan^4(x)+\cot^4(x)>0[/TEX] on the entire domain (reals that are not multiples of [TEX]\pi[/TEX]).

Both terms in [TEX]\tan^4(x)+\cot^4(x)[/TEX] are non-negative in the reals, so clearly the sum itself is non-negative.

[TEX]\tan^4(x)+\cot^4(x)[/TEX] cannot equal 0 because then either [TEX]\tan^4(x)=-\cot^4(x)[/TEX] (which per the line above is not possible) OR [TEX]\tan^4(x)=\cot^4(x)=0[/TEX], which also can't happen because [TEX]\cot^4(x)=\tfrac{1}{\tan^4(x)}[/TEX] and 0 cannot be a denominator.

As such, [TEX]\tan^4(x)+\cot^4(x)>0[/TEX].

And with this new information, [TEX]\sec^4(x) + \csc^4(x) = \underbrace{\left[\tan^4(x) + \cot^4(x)\right]}_{\text{always positive!}} + 2\underbrace{(\tan^2(x)+\cot^2(x))}_{\text{can show positive similarly}} + 2 > 2[/TEX].

It at long last follows that the range of [TEX]\sec^4(x) + \csc^4(x)[/TEX] is [TEX]\left(2,\infty\right)[/TEX].

If someone has an argument for a left bracket instead of my left open parenthesis, I'd love to hear it!

-Andy

#### MarkFL

Staff member
I checked first for the minima at W|A, and it suggests, that given:

$f(x)=\sec^4(x)+\csc^4(x)$

then the global minimum is $f(x)=8$, at:

$x=2(k\pi\pm\tan^{-1}(1\pm\sqrt{2}))$ where $k\in\mathbb{Z}$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi abender!

You are right that the range will not include zero.
The actual range is a subset of $[0,\infty)$, but also a subset of $(0,\infty)$.

Btw, the range is also not $(2,\infty)$.

What you have, is:

$\sec^4 x + \csc^4 x = \dfrac{1}{\cos^4 x} +\dfrac{1}{\sin^4 x}$

This expression is always positive, and it tends to infinity at every multiple of $\dfrac \pi 2$.
Since cosine and sine are symmetric, it will take its lowest value at $\dfrac \pi 4$, where it takes the value 8.
So the range is $[8, \infty)$.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
$x=2(k\pi\pm\tan^{-1}(1\pm\sqrt{2}))$ where $k\in\mathbb{Z}$
Wow!! That is a pretty complex way of W|A to say $x=\dfrac \pi 4 + k \dfrac \pi 2$.

#### MarkFL

Staff member
Wow!! That is a pretty complex way of W|M to say $x=\dfrac \pi 4 + k \dfrac \pi 2$.
It sure is!...and I didn't bother to try to simplify either...

I simply noted that the range was different than the OP suggested and was going to let them go "back to the drawing board."

#### Klaas van Aarsen

##### MHB Seeker
Staff member
It sure is!...and I didn't bother to try to simplify either...

I simply noted that the range was different than the OP suggested and was going to let them go "back to the drawing board."
Yes, I'm surprised that W|A did not give a simpler solution.
Either way, it's not easy to simplify.
I didn't know that $\arctan(\sqrt 2 - 1) = \dfrac \pi 8$.
And I wouldn't know how to figure it out in this direction (the other way around I'd use the double angle formulas).
So I learned something new today.

#### abender

##### New member
Yes, I'm surprised that W|A did not give a simpler solution.
Either way, it's not easy to simplify.
I didn't know that $\arctan(\sqrt 2 - 1) = \dfrac \pi 8$.
And I wouldn't know how to figure it out in this direction (the other way around I'd use the double angle formulas).
So I learned something new today.
You aren't kiddin. Wow.

#### abender

##### New member
I feel stupid. I responded with a very very very very very rough lower bound. Who does this? "... plus 2 times something positive..." Of course I then end up with a superset of the range. Go Ravens!

#### MarkFL

Staff member
Suppose we draw a right triangle, and with respect to one of the two acute angles (we'll call this angle $\theta$), we state:

$\tan(\theta)=\sqrt{2}-1$

and so:

$\displaystyle \sin(\theta)=\frac{\sqrt{2}-1}{\sqrt{2\sqrt{2}(\sqrt{2}-1)}}$

$\displaystyle \cos(\theta)=\frac{1}{\sqrt{2\sqrt{2}(\sqrt{2}-1)}}$

Now we may then state:

$\displaystyle \sin(2\theta)=2\sin(\theta)\cos(\theta)=\frac{2( \sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}=\frac{1}{\sqrt{2}}$

Hence:

$\displaystyle 2\theta=\frac{\pi}{4}$

$\displaystyle \theta=\frac{\pi}{8}$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
For fun I tried to find the common values for the trigonometric functions.
Wiki did not contain these specific values.
Now it does.

#### anemone

##### MHB POTW Director
Staff member
Find the range of $$\sec^4(x)+\csc^4(x)$$......

$\displaystyle \sec^4 x+\csc^4 x=\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}$

By AM-GM inequality, we have

$\displaystyle \frac{\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}}{2} \ge \sqrt{\frac{1}{\cos^4 x}.\frac{1}{\sin^4 x}}$

$\displaystyle \frac{\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x}}{2} \ge \frac{1}{\cos^2 x.\sin^2 x}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2}{\cos^2 x.\sin^2 x}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2}{(\cos x.\sin x)^2}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{2.2.2}{(2\cos x.\sin x)^2}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{8}{\sin^2 2x}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{8}{\frac{1-\cos 2x}{2}}$

$\displaystyle \frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge \frac{16}{1-\cos 2x}$

Since $\displaystyle 0 \le 1-\cos 2x \le 2$, we can conclude that $\displaystyle\frac{1}{\cos^4 x}+\frac{1}{\sin^4 x} \ge 8$, that is, the range of $\displaystyle \sec^4 x+\csc^4 x$ is [8, ∞).