Question

Is the following equation fully simplified?
Equation Below:

Answer 1

\[f(x)= \sqrt{(\sin^2(x)-2)^2}-\sqrt{(\sin^2(x)+1)^2}\]

Answer 2

hmmm what do you think?

Answer 3

I think it is because i can't just cancel the radical and the square (because of the possiblity that the inside is negative

Answer 4

I was puzzled because of something like this:
\[\sqrt{(-2)^2}\neq-2\]

Answer 5

well... hmm kinda, yes
but one could say \(\sqrt{(-2)^2}\to
\begin{cases}
-2\\
+2
\end{cases}\)

Answer 6

if you use the +/- part

but, for an expression, it depends on context
in this expression, I'd get rid of the radical and exponent though
and yes
you're correct, the radical IS a constraint on the function
but the question is, can it be simplified further

and I think it can, you can have a simplified version, keeping in mind the constraints

Answer 7

but if i complete that simplification, i get \[f(x)=-3\] which fails for all x

Answer 8

hmm

Answer 9

let's try one value
\(\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2}\)

what would that give?

Answer 10

-1

Answer 11

hmm -1?

Answer 12

ahh, i believe I've solved it. Since the first radicand will always be negative before it is squared, I will simplify take its absolute value, by flipping the terms, to get \[\sqrt{(\sin^2(x)-2)^2}=\sqrt{(2-\sin^2(x))^2}=2-\sin^2(x)\] from here, I can simplify the second radicand easily because it is always positive and thus is simply \[\sqrt{(\sin^2(x)+1)^2}=\sin^2(x)+1\] putting it all together, i get \[f(x)=2-\sin^2(x)-(\sin^2(x)+1)=1-2\sin^2(x)\] which equals, finally, \[\cos(2x)\] does that work?

Answer 13

hmmm the first radicand, actually both, are raised to the 2nd power, and thus, even if you get a negative value from the sine addition, the exponent will make it positive anyway

-* - = +

Answer 14

but with respect to eliminating the radical entirely, doesn't the flipping work?

Answer 15

ahemm nope

2-x \(\ne\) x-2

Answer 16

hmm actually... shoot got a mistake there.....

Answer 17

wheres the mistake?

Answer 18

hmm well. my -3 is... mistaken for one

Answer 19

hmmm

Answer 20

either way.. .you can't simply flip the terms in the radicand though, unless you multiply them by a -1

Answer 21

Look at this, attempted with x=pi/2\[\sqrt{(\sin^2(\pi/2)-2)^2}=1\] but also,\[\sqrt{(2-\sin^2(\pi/2))^2}=1\] So if i rewrite it in the second way, I can remove the radical.

Answer 22

hmmm
somehow... I do see the difference, expression wise

and yes, the 2nd power exponent, will make, whatever value inside, negative or not, positive

Answer 23

\(\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2}
\\ \quad \\

f\left( \frac{\pi }{2} \right)=\sqrt{[1^2-2]^2}-\sqrt{[1^2+1]^2}\implies f\left( \frac{\pi }{2} \right)=\sqrt{[-1]^2}-\sqrt{[2]^2}
\\ \quad \\
f\left( \frac{\pi }{2} \right)=-1-2\implies f\left( \frac{\pi }{2} \right)=-3
\\ \quad \\
f(0)=\sqrt{[sin^2(0)-2]^2}-\sqrt{[sin^2(0)+1]^2}
\\ \quad \\
f(0)=\sqrt{[0-2]^2}-\sqrt{[0+1]^2}\implies f(0)=-2-1\implies -3\)

so.... I"d think the -3 kinda checks out though

Answer 24

but yes, one could say, it CAN be simplified further
by getting rid of the radical and exponent
BUT
the simplified version will still maintain the constraints of the original one

Answer 25

and yes, I'm aware that \(\bf \sqrt{(-2)^2}\ne -2\qquad or\qquad \sqrt{(-1)^2}\ne -1\)

but all that, depends on the context at hand

Answer 26

I mean... the same can be said... say. of an inverse trigonometric function
their range is \(\textit{Inverse Trigonometric Identities}
\\ \quad \\

\begin{array}{cccl}

Function&{\color{brown}{ Domain}}&{\color{blue}{ Range}}\\
\hline\\
{\color{blue}{ y}}=sin^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1&-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\\ \quad \\
{\color{blue}{ y}}=cos^{-1}({\color{brown}{ \theta}})&-1\le {\color{brown}{ \theta}} \le 1& 0 \le {\color{blue}{ y}}\le \pi
\\ \quad \\
{\color{blue}{ y}}=tan^{-1}({\color{brown}{ \theta}})&-\infty\le {\color{brown}{ \theta}} \le +\infty &-\frac{\pi}{2}\le {\color{blue}{ y}}\le \frac{\pi}{2}
\end{array}\)

but depending on context, you can take the arcSine or arcCosine of a value
and get just a "reference angle", and use it on 2nd, 3rd, or 4th quadrants

even though the inverse function, may not extend to it

Answer 27

Wow, i thought there was just a solid rule explaining \[\sqrt{(-2)^2}\] I learned that a radical can never be negative no matter what the conditions inside it, so i don't know, I might just put cos(2x), because it checks out on all my tries both positive and negative

Answer 28

hehe
bear in mind that notations and formulas, are only phenomena representations

some results in math are called "math fallacies" or "extraneous"
because the procedure is correct, bu the answer is ODD

so, they're correct mathematically, but wrong logically, and the logic throws them out
in this case, is an extraneous case

is correct mathematically, logicaly it makes no much sense, you're correct
thus is "extraneous"

but depending on the context they're used, extraneous results can be valid or not

in this case, we don't have an explicit context or phenomena for this expression
thus the extraneous result is ok

Answer 29

ok, thank you