Is the following equation fully simplified?
Equation Below:

- Rizags

Is the following equation fully simplified?
Equation Below:

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- Rizags

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

- jdoe0001

hmmm what do you think?

- Rizags

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

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## More answers

- Rizags

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

- jdoe0001

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

- jdoe0001

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

- Rizags

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

- jdoe0001

hmm

- jdoe0001

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?

- Rizags

-1

- jdoe0001

hmm -1?

- Rizags

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?

- jdoe0001

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
-* - = +

- Rizags

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

- jdoe0001

ahemm nope
2-x \(\ne\) x-2

- jdoe0001

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

- Rizags

wheres the mistake?

- jdoe0001

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

- jdoe0001

hmmm

- jdoe0001

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

- Rizags

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.

- jdoe0001

hmmm
somehow... I do see the difference, expression wise
and yes, the 2nd power exponent, will make, whatever value inside, negative or not, positive

- jdoe0001

\(\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2}
\\ \quad \\
f\left( \frac{\pi }{2} \right)=\sqrt{[1^2-2]^2}-\sqrt{[1^2+1]^2}\implies f\left( \frac{\pi }{2} \right)=\sqrt{[-1]^2}-\sqrt{[2]^2}
\\ \quad \\
f\left( \frac{\pi }{2} \right)=-1-2\implies f\left( \frac{\pi }{2} \right)=-3
\\ \quad \\
f(0)=\sqrt{[sin^2(0)-2]^2}-\sqrt{[sin^2(0)+1]^2}
\\ \quad \\
f(0)=\sqrt{[0-2]^2}-\sqrt{[0+1]^2}\implies f(0)=-2-1\implies -3\)
so.... I"d think the -3 kinda checks out though

- jdoe0001

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

- jdoe0001

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

- jdoe0001

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

- Rizags

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

- jdoe0001

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

- Rizags

ok, thank you

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