# Drawing with trigonometric function

• l46kok
Rotation Matrix for 45 Degrees:I can just model the equation with sines. So y_1 = 300 * Sin(pi*x / 1000) (Assuming 300 is the highest point in the curve)y_2 = (1000/pi) * sin^-1(x/300)

## Homework Statement

Come up with a set of equations to draw the following picture:

http://img197.imageshack.us/img197/7428/equation.jpg [Broken]

## The Attempt at a Solution

The circle is easy.

1000 = sqrt(x^2+y^2)

I can just model the equation with sines. So

y_1 = 300 * Sin(pi * x / 1000) (Assuming 300 is the highest point in the curve)

The sine graph on the y-axis can be easily tilted by switching x and y. Therefore:

x = 300 * Sin(pi*y_2 / 1000)

sin^-1 (x/300) = pi * y_2 / 1000

y_2 = (1000/pi) * sin^-1(x/300)

Is my y_2 correct?

And I'm kinda stumbling on how to translate the axis in 45 degree angle to draw the sine graph like that. Any help would be appreciated.

Last edited by a moderator:
l46kok said:
I can just model the equation with sines. So

y_1 = 300 * Sin(pi * x / 1000) (Assuming 300 is the highest point in the curve)
You should probably add a restriction for x, like -1000 ≤ x ≤ 1000.

l46kok said:
The sine graph on the y-axis can be easily tilted by switching x and y. Therefore:

x = 300 * Sin(pi*y_2 / 1000)

sin^-1 (x/300) = pi * y_2 / 1000

y_2 = (1000/pi) * sin^-1(x/300)

Is my y_2 correct?
Technically, no, because x has the domain restriction -π/2 ≤ x ≤ π/2. Must all equations be y as functions of x? If not, I would just leave it in terms of x. Also, I think you need a negative in front of the R.H.S.:
x = -300 * Sin(pi*y_2 / 1000)

No it can be in terms of X too.

How should I modelize the lines which goes at 45 degree angle?

l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)

l46kok said:
No it can be in terms of X too.

How should I modelize the lines which goes at 45 degree angle?
Are you familiar with rotation of axes? You'll need to make use of it, I think.

eumyang said:
No it's not,

l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)

:tongue:

eumyang said:

Are you familiar with rotation of axes? You'll need to make use of it, I think.

Yeah, translating the axes by defining a new set of axes such as x' and y', then rotating it via trigonometric function but I was struggling big time about that. Any suggestions?

Mentallic said:
No it's not,
:tongue:
Wow... that's was embarrassing. Let's try that again.
l46kok said:
The circle is easy.

1000 = sqrt(x^2+y^2)
This only gives you the upper semicircle. Define this implicitly instead:
x2+y2=10002

l46kok said:
Yeah, translating the axes by defining a new set of axes such as x' and y', then rotating it via trigonometric function but I was struggling big time about that. Any suggestions?
Do you know the equations that define x' and y' in terms of x, y, and θ? Take the equation
y' = 300 * sin(pi*x'/1000)
and replace with what x' and y' equals. Then find a new set of equations for x' and y', in terms of a different angle θ, and replace into the equation above.

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Call your original function y = f(x). The easiest way to rotate is to use the rotation matrix:

$$R(x) = \left[ \begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right] \left[ \begin{array}{c} x\\ f(x) \end{array}\right]$$

Pick any θ you wish and plot it parametrically for x from 0 to 1000.