anonymous
  • anonymous
Can someone help me simplify this? y=3cos^3 (arcsin(cuberoot(x/3)))
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[y=3\cos^3(\sin^{-1} \sqrt[3]{x/3})\]
anonymous
  • anonymous
@ganeshie8 @Data_LG2
anonymous
  • anonymous
WabbitStudio Z80 Software Tools - Home https://wabbit.codeplex.com/

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anonymous
  • anonymous
@SithsAndGiggles
geerky42
  • geerky42
*
anonymous
  • anonymous
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.
anonymous
  • anonymous
It has to simplify to y = ?
anonymous
  • anonymous
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 \(3\).
anonymous
  • anonymous
This is the second half of eliminating the parameter of a parametric equation problem. So It can't end at y = 3cos^3 (t)
anonymous
  • anonymous
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 \(\cos t\)?
anonymous
  • anonymous
In other words, what's \(\dfrac{\color{red}?}{3}\)? Use the Pythagorean theorem to find \(\color{red}?\) first.
anonymous
  • anonymous
I dont know how the \[\sqrt[3]{x}\] would be squared and what that would come out to
anonymous
  • anonymous
Well, not everything has the luxury of simplifying nicely. \[\begin{align*} 3^2&=\left(\sqrt[3]x\right)^2+\color{red}?^2\\ 9-x^{2/3}&=\color{red}?^2&\text{since }\sqrt[n]x=x^{1/n}\text{ and }(x^a)^b=x^{ab}\\ \color{red}?&=\sqrt{9-x^{2/3}} \end{align*}\]
anonymous
  • anonymous
well ? would be \[\sqrt{9-x ^{2/3}}\]
anonymous
  • anonymous
since we can't find x
anonymous
  • anonymous
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?
anonymous
  • anonymous
y
anonymous
  • anonymous
True, but not the right answer. \[\text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2.\\ \text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^3t=\cdots\]
anonymous
  • anonymous
oh. that cubed (to lazy to type it)
anonymous
  • anonymous
@hero can you finish off what he started?
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
what's your question?
anonymous
  • anonymous
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.
jim_thompson5910
  • jim_thompson5910
Can you simplify the cos^2 equation? hint: \(\Large \sqrt{x^2} = (\sqrt{x})^2 = x\) where x is nonnegative
anonymous
  • anonymous
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.
anonymous
  • anonymous
cos^3?
jim_thompson5910
  • jim_thompson5910
Can you simplify this \[\Large \cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2\]
anonymous
  • anonymous
\[\frac{ 9-x ^{2/3} }{ 9 }\]
jim_thompson5910
  • jim_thompson5910
very good
jim_thompson5910
  • jim_thompson5910
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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