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anonymous

  • one year ago

Can someone help me simplify this? y=3cos^3 (arcsin(cuberoot(x/3)))

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  1. anonymous
    • one year ago
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    \[y=3\cos^3(\sin^{-1} \sqrt[3]{x/3})\]

  2. anonymous
    • one year ago
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    @ganeshie8 @Data_LG2

  3. anonymous
    • one year ago
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    WabbitStudio Z80 Software Tools - Home https://wabbit.codeplex.com/

  4. anonymous
    • one year ago
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    @SithsAndGiggles

  5. geerky42
    • one year ago
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    *

  6. anonymous
    • one year ago
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    Let \(t=\arcsin\dfrac{\sqrt[3]x}{3}\), so that \(\sin t=\dfrac{\sqrt[3]x}{3}\). You have enough info to determine \(\cos t\) from here.

  7. anonymous
    • one year ago
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    It has to simplify to y = ?

  8. anonymous
    • one year ago
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    Recall how sine is defined in terms of a triangle's sides: \(\sin(\text{some angle})=\dfrac{\text{length of side opposite the angle}}{\text{length of hypotenuse}}\). So if I told you that \(\sin\theta=\dfrac{1}{2}\), for instance, you could draw up a right triangle that satisfies this to use as a reference. |dw:1435441331195:dw| What's the length of the missing side? What's \(\cos\theta\)? You can use the same principle for your problem to determine \(\cos t\). The actual "simplification" from that point is just a matter of raising \(\cos t\) to the third power and multiplying by \(3\).

  9. anonymous
    • one year ago
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    This is the second half of eliminating the parameter of a parametric equation problem. So It can't end at y = 3cos^3 (t)

  10. anonymous
    • one year ago
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    And indeed it doesn't! The problem here is to find an equivalent expression for \(\cos\left(\arcsin\cdots\right)\). The substitution is only used to make it easier to view the \(\arcsin\) component as an angle. In your problem, |dw:1435441649836:dw| So what's \(\cos t\)?

  11. anonymous
    • one year ago
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    In other words, what's \(\dfrac{\color{red}?}{3}\)? Use the Pythagorean theorem to find \(\color{red}?\) first.

  12. anonymous
    • one year ago
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    I dont know how the \[\sqrt[3]{x}\] would be squared and what that would come out to

  13. anonymous
    • one year ago
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    Well, not everything has the luxury of simplifying nicely. \[\begin{align*} 3^2&=\left(\sqrt[3]x\right)^2+\color{red}?^2\\ 9-x^{2/3}&=\color{red}?^2&\text{since }\sqrt[n]x=x^{1/n}\text{ and }(x^a)^b=x^{ab}\\ \color{red}?&=\sqrt{9-x^{2/3}} \end{align*}\]

  14. anonymous
    • one year ago
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    well ? would be \[\sqrt{9-x ^{2/3}}\]

  15. anonymous
    • one year ago
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    since we can't find x

  16. anonymous
    • one year ago
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    We're not trying to find \(x\), we're just rewriting \(\cos\left(\arcsin\dfrac{\sqrt[3]x}{3}\right)\) as a simpler expression containing \(x\). |dw:1435442613564:dw| This tells us that \(\cos\left(\arcsin\dfrac{\sqrt[3]x}{3}\right)=\cos t=\dfrac{\sqrt{9-x^{2/3}}}{3}\). So what is \(3\cos^3t\) equivalent to?

  17. anonymous
    • one year ago
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    y

  18. anonymous
    • one year ago
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    True, but not the right answer. \[\text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2.\\ \text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^3t=\cdots\]

  19. anonymous
    • one year ago
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    oh. that cubed (to lazy to type it)

  20. anonymous
    • one year ago
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    @hero can you finish off what he started?

  21. anonymous
    • one year ago
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    @jim_thompson5910

  22. jim_thompson5910
    • one year ago
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    what's your question?

  23. anonymous
    • one year ago
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    The equation at the top needs to be simplified. He stopped halfway through (I believe) and I've been stuck on this question for 2 hours trying to figure out how to do this crap.

  24. jim_thompson5910
    • one year ago
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    Can you simplify the cos^2 equation? hint: \(\Large \sqrt{x^2} = (\sqrt{x})^2 = x\) where x is nonnegative

  25. anonymous
    • one year ago
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    I've read through my text book 3 times trying to figure it out and the problems for the homework are much harder than my example problems in the lesson.

  26. anonymous
    • one year ago
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    cos^3?

  27. jim_thompson5910
    • one year ago
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    Can you simplify this \[\Large \cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2\]

  28. anonymous
    • one year ago
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    \[\frac{ 9-x ^{2/3} }{ 9 }\]

  29. jim_thompson5910
    • one year ago
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    very good

  30. jim_thompson5910
    • one year ago
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    so... \[\Large \cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2\] \[\Large \cos^2t=\frac{9-x^{2/3}}{9}\] \[\Large \cos t * \cos^2t=\cos t * \frac{9-x^{2/3}}{9}\] \[\Large \cos^3t=\frac{\sqrt{9-x^{2/3}}}{3} * \frac{9-x^{2/3}}{9}\] \[\Large \cos^3t=\frac{\left(9-x^{2/3}\right)\sqrt{9-x^{2/3}}}{27}\]

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