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Rizags

  • one year ago

Is the following equation fully simplified? Equation Below:

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  1. Rizags
    • one year ago
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    \[f(x)= \sqrt{(\sin^2(x)-2)^2}-\sqrt{(\sin^2(x)+1)^2}\]

  2. jdoe0001
    • one year ago
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    hmmm what do you think?

  3. Rizags
    • one year ago
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    I think it is because i can't just cancel the radical and the square (because of the possiblity that the inside is negative

  4. Rizags
    • one year ago
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    I was puzzled because of something like this: \[\sqrt{(-2)^2}\neq-2\]

  5. jdoe0001
    • one year ago
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    well... hmm kinda, yes but one could say \(\sqrt{(-2)^2}\to \begin{cases} -2\\ +2 \end{cases}\)

  6. jdoe0001
    • one year ago
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    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

  7. Rizags
    • one year ago
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    but if i complete that simplification, i get \[f(x)=-3\] which fails for all x

  8. jdoe0001
    • one year ago
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    hmm

  9. jdoe0001
    • one year ago
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    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?

  10. Rizags
    • one year ago
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    -1

  11. jdoe0001
    • one year ago
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    hmm -1?

  12. Rizags
    • one year ago
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    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?

  13. jdoe0001
    • one year ago
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    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 -* - = +

  14. Rizags
    • one year ago
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    but with respect to eliminating the radical entirely, doesn't the flipping work?

  15. jdoe0001
    • one year ago
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    ahemm nope 2-x \(\ne\) x-2

  16. jdoe0001
    • one year ago
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    hmm actually... shoot got a mistake there.....

  17. Rizags
    • one year ago
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    wheres the mistake?

  18. jdoe0001
    • one year ago
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    hmm well. my -3 is... mistaken for one

  19. jdoe0001
    • one year ago
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    hmmm

  20. jdoe0001
    • one year ago
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    either way.. .you can't simply flip the terms in the radicand though, unless you multiply them by a -1

  21. Rizags
    • one year ago
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    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.

  22. jdoe0001
    • one year ago
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    hmmm somehow... I do see the difference, expression wise and yes, the 2nd power exponent, will make, whatever value inside, negative or not, positive

  23. jdoe0001
    • one year ago
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    \(\bf f\left( \frac{\pi }{2} \right)= \sqrt{[\sin^2\left( \frac{\pi }{2} \right)-2]^2}-\sqrt{[\sin^2\left( \frac{\pi }{2} \right)+1]^2} \\ \quad \\ f\left( \frac{\pi }{2} \right)=\sqrt{[1^2-2]^2}-\sqrt{[1^2+1]^2}\implies f\left( \frac{\pi }{2} \right)=\sqrt{[-1]^2}-\sqrt{[2]^2} \\ \quad \\ f\left( \frac{\pi }{2} \right)=-1-2\implies f\left( \frac{\pi }{2} \right)=-3 \\ \quad \\ f(0)=\sqrt{[sin^2(0)-2]^2}-\sqrt{[sin^2(0)+1]^2} \\ \quad \\ f(0)=\sqrt{[0-2]^2}-\sqrt{[0+1]^2}\implies f(0)=-2-1\implies -3\) so.... I"d think the -3 kinda checks out though

  24. jdoe0001
    • one year ago
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    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

  25. jdoe0001
    • one year ago
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    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

  26. jdoe0001
    • one year ago
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    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

  27. Rizags
    • one year ago
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    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

  28. jdoe0001
    • one year ago
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    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

  29. Rizags
    • one year ago
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    ok, thank you

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