Question

HELP I can't figure out how to eliminate the parameters.
x = 3 sin^3t
y = 3 cos^3t

Answer 1

are you aware of the identity: \(\cos^2(\theta)+\sin^2(\theta)=1\) ?

Answer 2

Yes, but I have no idea how to get to that.

Answer 3

try squaring each of your expressions and then adding them :)

Answer 4

I'm sorry, but what?

Answer 5

@asnaseer , the sin/cos are cubed functions

Answer 6

sorry - I may have misread your question, I read it as \(x=3\sin(3t)\)

Answer 7

No, it is read as \[3 \sin^3 t \]

Answer 8

so are your equations:\[x=3\sin^3(t)\]\[y=3\cos^3(t)\]

Answer 9

yes

Answer 10

:)

Answer 11

\[\sin t = (\frac{x}{3})^{1/3}\]
\[y = 3 \cos t (1-\sin^2 t)\]
make the substitution

Answer 12

I have little to no idea as to how to make it squared and not cubed.

Answer 13

sorry
\[\cos t = (1-\sin^2 t)^{1/2}\]

Answer 14

you can rearrange your equations to become:\[\sin(t)=\sqrt[3]{x/3}\]\[\cos(t)=\sqrt[3]{y/3}\]and then use the cos/sin identity I gave above

Answer 15

Shouldn't Sine and Cosine be squared to work in the Pythagorean theorem identity?

Answer 16

Use either the response given by @dumbcow or use the fact that:\[\cos^2(t)+\sin^2(t)=1\]

Answer 17

they are both equivalent

Answer 18

I know this is redundant, but Sine and Cosine aren't supposed to be squared?

Answer 19

do you agree that for any angle \(\theta\):\[\cos^2(\theta)+\sin^2(\theta)=1\]

Answer 20

yes

Answer 21

So just let \(\theta=t\) and we get:\[\cos^2(t)+\sin^2(t)=1\]

Answer 22

Now all you need to do is to substitute the expressions for \(\cos(t)\) and \(\sin(t)\) that we derived above and you are done :)