1. ## Cartesian equations

Given $\displaystyle x=2 \sin (nt+\frac{\pi}{3})$ and $\displaystyle y=4 \sin (nt+ \frac{\pi}{6})$, express $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$. Find the Cartesian equation of the locus of the point $\displaystyle (x,y)$ as $\displaystyle t$ varies.

I have expressed $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$ already.
$\displaystyle x=2 \sin (nt+\frac{\pi}{3})$
$\displaystyle x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$\displaystyle x= \sin nt+ \sqrt{3} \cos nt$
and
$\displaystyle y=4 \sin (nt+ \frac{\pi}{6})$
$\displaystyle y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$\displaystyle y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks

2. Originally Posted by arze
Given $\displaystyle x=2 \sin (nt+\frac{\pi}{3})$ and $\displaystyle y=4 \sin (nt+ \frac{\pi}{6})$, express $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$. Find the Cartesian equation of the locus of the point $\displaystyle (x,y)$ as $\displaystyle t$ varies.

I have expressed $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$ already.
$\displaystyle x=2 \sin (nt+\frac{\pi}{3})$
$\displaystyle x=2(\frac{1}{2} \sin nt+\sqrt{3} \cos nt)$
$\displaystyle x= \sin nt+ \sqrt{3} \cos nt$
and
$\displaystyle y=4 \sin (nt+ \frac{\pi}{6})$
$\displaystyle y=4(\frac{\sqrt{3}}{2} \cos nt+\frac{1}{2} \sin nt)$
$\displaystyle y= 2\sqrt{3} \cos nt +2 \sin nt$
Now my problem is how do i form the Cartesian equation in x and y as t varies?
I take n as constant and t varies? Can anyone explain?
Thanks
$\displaystyle x=2 \sin (nt+\frac{\pi}{3})$ and $\displaystyle y=4 \sin (nt+ \frac{\pi}{6})$

$\displaystyle x=2 \sin (nt+\frac{\pi}{6}+\frac{\pi}{6})$

let $\displaystyle \alpha=nt+\frac{\pi}{6}$

$\displaystyle x=2 \sin (\alpha+\frac{\pi}{6})$ and $\displaystyle y=4 \sin (\alpha)$

$\displaystyle \frac{x}{2}=\sin (\alpha)\cos(\frac{\pi}{6}) + \cos (\alpha)\sin(\frac{\pi}{6})$

and

$\displaystyle \frac{y}{4}=\sin (\alpha)$

use $\displaystyle \sin^2 (\alpha) + \cos^2(\alpha) = 1$ to eliminate $\displaystyle \alpha$ and get your equation

3. Hello, arze!

Given: .$\displaystyle \begin{array}{ccc}x &=&2 \sin \left(nt+\frac{\pi}{3}\right) \\ y &=& 4 \sin \left(nt+ \frac{\pi}{6}\right)\end{array}$

Express $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$.

Find the Cartesian equation of the locus of the point $\displaystyle (x,y)$ as $\displaystyle t$ varies.

I have expressed $\displaystyle x$ and $\displaystyle y$ in terms of $\displaystyle \sin nt$ and $\displaystyle \cos nt$ already.

. . $\displaystyle \begin{array}{ccccc}x\:=\: 2\sin\left(nt+\tfrac{\pi}{3}\right) \:=\:2\left(\tfrac{1}{2}\sin nt+\tfrac{\sqrt{3}}{2}\cos nt\right) &\Rightarrow& x\:=\: \sin nt+ \sqrt{3} \cos nt \\ \\[-4mm] y\:=\:4\sin\left(nt+ \tfrac{\pi}{6}\right) \:=\:4\left(\tfrac{\sqrt{3}}{2} \cos nt+\tfrac{1}{2} \sin nt\right) &\Rightarrow& y \:=\: 2\sqrt{3} \cos nt +2 \sin nt \end{array}$

. . . . . Good work!

Now my problem is how do i form the Cartesian equation in $\displaystyle x$ and $\displaystyle y$ as $\displaystyle t$ varies?

We have: .$\displaystyle \begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$

$\displaystyle \begin{array}{ccccc}\text{Square }{\color{blue}(1)}: & x^2 &=& \sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + 3\cos^2\!nt \\ \text{Square }{\color{blue}(2)}: & \dfrac{y^2}{4} &=& 3\sin^2\!nt + 2\sqrt{3}\sin nt\cos nt + \cos^2\!nt \end{array}$

Add: .$\displaystyle x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$

. . . . $\displaystyle x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$

. . . . $\displaystyle x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$

4. Originally Posted by Soroban
Hello, arze!

We have: .$\displaystyle \begin{array}{cccc}x &=& \sin nt + \sqrt{3}\cos nt & {\color{blue}(1)} \\ \\[-4mm] \dfrac{y}{2} &=& \sqrt{3}\sin nt + \cos nt & {\color{blue}(2)}\end{array}$

Add: .$\displaystyle x^2 + \frac{y^2}{4} \;=\;4\sin^2\!nt + 4\sqrt{3}\sin nt\cos nt + 4\cos^2\!nt$

. . . . $\displaystyle x^2 + \frac{y^2}{4} \;=\;4\underbrace{\left(\sin^2\!nt + \cos^2\!nt\right)}_{\text{This is 1}} + 2\sqrt{3}\underbrace{\left(2\sin nt\cos nt\right)}_{\text{This is }\sin2nt}$

. . . . $\displaystyle x^2 + \frac{y^2}{4} \;=\;4 + 2\sqrt{3}\sin(2nt)$
if you want to do it that way, solve (1) and (2) for sin and cos:

$\displaystyle 2\sin nt = \frac{\sqrt3y}{2}-x$

$\displaystyle 2\cos nt = \sqrt3x-\frac{y}{2}$

then use

$\displaystyle \sin^2nt + \cos^2nt = 1$

to get an equation in x and y

5. Hello, Tesla!

Absolutely right!

It's an old habit.

Whenever I see something like: .$\displaystyle \begin{array}{ccc}x &=& a\sin\theta \\ y &=& b\cos\theta\end{array}$

. . I tend to use the square-and-add approach to simplify.

I must learn to get out of my "box".