Rizags
  • Rizags
Is the following equation fully simplified? Equation Below:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Rizags
  • Rizags
\[f(x)= \sqrt{(\sin^2(x)-2)^2}-\sqrt{(\sin^2(x)+1)^2}\]
jdoe0001
  • jdoe0001
hmmm what do you think?
Rizags
  • 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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Rizags
  • Rizags
I was puzzled because of something like this: \[\sqrt{(-2)^2}\neq-2\]
jdoe0001
  • jdoe0001
well... hmm kinda, yes but one could say \(\sqrt{(-2)^2}\to \begin{cases} -2\\ +2 \end{cases}\)
jdoe0001
  • 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
  • Rizags
but if i complete that simplification, i get \[f(x)=-3\] which fails for all x
jdoe0001
  • jdoe0001
hmm
jdoe0001
  • 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
  • Rizags
-1
jdoe0001
  • jdoe0001
hmm -1?
Rizags
  • 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
  • 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
  • Rizags
but with respect to eliminating the radical entirely, doesn't the flipping work?
jdoe0001
  • jdoe0001
ahemm nope 2-x \(\ne\) x-2
jdoe0001
  • jdoe0001
hmm actually... shoot got a mistake there.....
Rizags
  • Rizags
wheres the mistake?
jdoe0001
  • jdoe0001
hmm well. my -3 is... mistaken for one
jdoe0001
  • jdoe0001
hmmm
jdoe0001
  • jdoe0001
either way.. .you can't simply flip the terms in the radicand though, unless you multiply them by a -1
Rizags
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Rizags
ok, thank you

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