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

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

Mathematics
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\[y=3\cos^3(\sin^{-1} \sqrt[3]{x/3})\]
WabbitStudio Z80 Software Tools - Home https://wabbit.codeplex.com/

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