# Thread: Trig. - Elimination parameters

1. ## Trig. - Elimination parameters

Hey guys

How do i eliminate x in the following case: x = a sinx and y = b tan(x/2)?

I could do the basic questions but I have no clue how to do this one.

Thanx

2. Originally Posted by xwrathbringerx
Hey guys

How do i eliminate x in the following case: x = a sinx and y = b tan(x/2)?

I could do the basic questions but I have no clue how to do this one.

Thanx
I assume that x as an angle is not meant to be the same as x as a coordinate??

$x = a \sin t = 2a \sin \left( \frac{t}{2}\right) \cos \left( \frac{t}{2}\right)$ .... (1)

$y = b \tan \left( \frac{t}{2}\right) \Rightarrow \frac{y}{b} = \left( \frac{t}{2}\right)$ .... (2)

From equation (2): $\sin \left( \frac{t}{2} \right) = \frac{y}{\sqrt{y^2 + b^2}}$ and $\cos \left( \frac{t}{2}\right) = \frac{b}{\sqrt{y^2 + b^2}}$

assuming $0 \leq t \leq \pi$ (why?)

Substitute these two expressions into equation (1).

3. Hello, xwrathbringerx!

Eliminate the parameter $\theta$: . $\begin{array}{cccc}x &=& a\sin\theta & {\color{red}(a)} \\ y &=& b\tan\frac{\theta}{2} & {\color{red}(b)} \end{array}$

From (b) we have: . $y \;=\;b\,\frac{\sin\frac{\theta}{2}}{\cos\frac{\the ta}{2}} \quad\Rightarrow\quad y\;=\;b\sqrt{\frac{1-\cos x}{1 + \cos x}}$

Square both sides: . $y^2 \;=\;b^2\,\frac{1-\cos x}{1+\cos x} \quad\Rightarrow\quad y^2 + y^2\cos x \;=\;b^2 - b^2\cos x$

. . $b^2\cos x + y^2\cos x \;=\;b^2-y^2 \quad\Rightarrow\quad (b^2+y^2)\cos x \;=\;b^2-y^2$

. . $\cos x \;=\;\frac{b^2-y^2}{b^2+y^2}$ .[1]

From (a), we have: . $x \;=\;a\sin x \quad\Rightarrow\quad x\;=\;a\sqrt{1-\cos^2\!x}$ .[2]

Substitute [1] into [2]: . $x \;=\;a\sqrt{1-\left(\frac{b^2-y^2}{b^2+y^2}\right)^2} \;=\;a\sqrt{\frac{(b^2+y^2)^2 - (b^2-y^2)^2}{(b^2+y^2)^2}}$

. . which simplifies to: . $x \;=\;a\sqrt{\frac{4b^2y^2}{(b^2+y^2)^2}} \;=\;a\,\frac{2by}{b^2+y^2}$

Therefore: . $x \;=\;\frac{2ab\,y}{b^2+y^2}$