## anonymous one year ago Can someone help me simplify this? y=3cos^3 (arcsin(cuberoot(x/3)))

1. anonymous

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

2. anonymous

@ganeshie8 @Data_LG2

3. anonymous

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4. anonymous

@SithsAndGiggles

5. geerky42

*

6. 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.

7. anonymous

It has to simplify to y = ?

8. 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 $$3$$.

9. 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)

10. 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 $$\cos t$$?

11. anonymous

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

12. anonymous

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

13. 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*}

14. anonymous

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

15. anonymous

since we can't find x

16. 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?

17. anonymous

y

18. anonymous

True, but not the right answer. $\text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^2t=\left(\frac{\sqrt{9-x^{2/3}}}{3}\right)^2.\\ \text{If }\cos t=\frac{\sqrt{9-x^{2/3}}}{3},\text{ then }\cos^3t=\cdots$

19. anonymous

oh. that cubed (to lazy to type it)

20. anonymous

@hero can you finish off what he started?

21. anonymous

@jim_thompson5910

22. jim_thompson5910

23. 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.

24. jim_thompson5910

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

25. 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.

26. anonymous

cos^3?

27. jim_thompson5910

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

28. anonymous

$\frac{ 9-x ^{2/3} }{ 9 }$

29. jim_thompson5910

very good

30. jim_thompson5910

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