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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

\[y=3\cos^3(\sin^{-1} \sqrt[3]{x/3})\]

- anonymous

@ganeshie8 @Data_LG2

- anonymous

WabbitStudio Z80 Software Tools - Home https://wabbit.codeplex.com/

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@SithsAndGiggles

- geerky42

*

- 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

It has to simplify to y = ?

- 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

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

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

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

- anonymous

I dont know how the \[\sqrt[3]{x}\] would be squared and what that would come out to

- 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

well ? would be \[\sqrt{9-x ^{2/3}}\]

- anonymous

since we can't find x

- 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

y

- 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

oh. that cubed (to lazy to type it)

- anonymous

@hero can you finish off what he started?

- anonymous

@jim_thompson5910

- jim_thompson5910

what's your question?

- 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

Can you simplify the cos^2 equation?
hint: \(\Large \sqrt{x^2} = (\sqrt{x})^2 = x\) where x is nonnegative

- 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

cos^3?

- jim_thompson5910

Can you simplify this
\[\Large \cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2\]

- anonymous

\[\frac{ 9-x ^{2/3} }{ 9 }\]

- jim_thompson5910

very good

- jim_thompson5910

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.