1. ## changing expression

Hi,

I was just wondering if you have an expression like this $sec^{4} \theta - tan^{4} \theta$ and if I take the sqaure root of this expression would it end up looking like this ; $sec^{2} \theta - tan^{2} \theta$?

thanks!

2. Originally Posted by Tweety
Hi,

I was just wondering if you have an expression like this $sec^{4} \theta - tan^{4} \theta$ and if I take the sqaure root of this expression would it end up looking like this ; $sec^{2} \theta - tan^{2} \theta$?

thanks!
in general, no. $\sqrt{a^4 - b^4} \ne a^2 - b^2$

but you can do this ....

$\sec^4{t} - \tan^4{t} =$

$(\sec^2{t} - \tan^2{t})(\sec^2{t} + \tan^2{t}) =$

$[(1 + \tan^2{t}) - \tan^2{t}](\sec^2{t} + \tan^2{t}) =$

$(1)(\sec^2{t} + \tan^2{t}) = \sec^2{t}+\tan^2{t} = 1 + 2\tan^2{t}$

3. Originally Posted by skeeter
in general, no. $\sqrt{a^4 - b^4} \ne a^2 - b^2$

but you can do this ....

$\sec^4{t} - \tan^4{t} =$

$(\sec^2{t} - \tan^2{t})(\sec^2{t} + \tan^2{t}) =$

$[(1 + \tan^2{t}) - \tan^2{t}](\sec^2{t} + \tan^2{t}) =$

$(1)(\sec^2{t} + \tan^2{t}) = \sec^2{t}+\tan^2{t} = 1 + 2\tan^2{t}$
Okay thank you. Also how do you know that this expression is a difference of two squares?

or is it in general any expression like that can be expressed as a difference of two squares?

4. Originally Posted by Tweety
Okay thank you. Also how do you know that this expression is a difference of two squares?

or is it in general any expression like that can be expressed as a difference of two squares?
If there are only two terms and both are squares it's likely to be the difference of two squares. Essentially keep a lookout for quadratics with no x term.

$a^n - b^n = (a^{\frac{n}{2}} - b^{\frac{n}{2}})(a^{\frac{n}{2}} + b^{\frac{n}{2}})$

This reduces to an integer for even values of n

In the world of trig powers $sin^n(x) = (sin(x))^n \: , \: n \in \mathbb{Z^+}$