# Range

• Feb 3rd 2013, 05:37 AM
kjchauhan
Range
Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$.

• Feb 3rd 2013, 06:19 AM
Plato
Re: Range
Quote:

Originally Posted by kjchauhan
Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$.

What have you done?

Have you at least graphed this expression?
• Feb 3rd 2013, 10:05 AM
abender
Re: Range
Quote:

Originally Posted by kjchauhan
Please help me to find the range of $sec^{4}(x)+cosec^{4}(x)$.

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

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

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

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
• Feb 3rd 2013, 10:46 AM
Plato
Re: Range
Quote:

Originally Posted by abender
[TEX]\sec^4(x) = (\sec^2(x))^2 =
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.
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!

Take a look at the graph of the function.
• Feb 3rd 2013, 11:21 AM
abender
Re: Range
$[8,\infty)$, evidently. My way was a fun adventure; I wanted to do it using identities for some bizarre reason. These problems nearly always boil down to the rudimentary sines and cosines. I see the airtight solution now, the symmetry, no plus 2 times "something positive". In retrospect, why the hell did I accept 2 as my best lower bound? Oh well, my Ravens are in the Super Bowl! Usually never post if I'm uncertain.