MHB What Is the Range of sec^4(x) + csc^4(x)?

[abender]

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:

Please help me to find the range of sec^{4}(x)+cosec^{4}(x).

Thanks in advance.

\sec^4(x) = (\sec^2(x))^2 = (1+\tan^2(x))^2 = \tan^4(x) + 2\tan^2 + 1

\csc^4(x) = (\csc^2(x))^2 = (1+\cot^2(x))^2 = \cot^4(x) + 2\cot^2 + 1\sec^4(x) + \csc^4(x) = \tan^4(x) + 2\tan^2 + 1 + \cot^4(x) + 2\cot^2 + 1= \tan^4(x) + \cot^4(x) + 2(\tan^2(x)+\cot^2(x)) + 2

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

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

BELOW is where others may disagree with me:

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

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

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

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

As such, \tan^4(x)+\cot^4(x)>0.

And with this new information, \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.

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

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

-Andy

 

[MarkFL]

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 $k\in\mathbb{Z}$

 

[I like Serena]

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 $[8, \infty)$.

 

[I like Serena]

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

Wow! That is a pretty complex way of W|A to say $x=\dfrac \pi 4 + k \dfrac \pi 2$. (Giggle)

 

[MarkFL]

Wow! That is a pretty complex way of W|M to say $x=\dfrac \pi 4 + k \dfrac \pi 2$. (Giggle)

It sure is!...and I didn't bother to try to simplify either...(Tmi)

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

 

[I like Serena]

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]

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]

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]

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:

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

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

 

[I like Serena]

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

 

[anemone]

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, ∞).