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B18JDM · Oct 29, 2010, 01:27 PM

I'm not sure if I'm doing this right so far I found r=0 theta=0 r=5 theta 45 r=0 theta 90 so on up to 675. How do I plot this and find the tips on an 8 pedal rose?

ebaines · Oct 29, 2010, 02:53 PM

Not quite right. Since $\sin \alpha = 0$ for $\alpha = 0,\ \pi,\ \2\pi$ , etc. then for the equation $R = 5 \sin 4 \theta$ , R=0 when:

$ \theta = 0, \\ \theta =\frac {\pi} 4 , \\ \theta =\frac {\pi} 2, \\ \theta = \frac {3\pi} 4 $
etc.

And R = +5 for: $\theta = \frac {\pi}8, \ \frac {5 \pi} 8$ , etc. and -5 for: $\theta = \frac {3 \pi} 8, \ \frac {7 \pi} 8$ ,etc.

Can you plot this out now?

B18JDM · Oct 31, 2010, 09:16 AM

Why is it pi/8 5pi/8... how did you formulate these angles?

Unknown008 · Oct 31, 2010, 09:37 AM

If R = 5, you get:

$5\sin4\theta = 5$

$\sin4\theta = 1$

$\alpha = \sin^{-1}(1)$

And $\alpha = \frac{\pi}{2}, \frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2},\frac{9\pi}{2}, \frac{11\pi}{2},\frac{13\pi}{2}, \frac{15\pi}{2}$

So,

$4\theta = \frac{\pi}{2}, \frac{3\pi}{2},\frac{5\pi}{2}, \frac{7\pi}{2},\frac{9\pi}{2}, \frac{11\pi}{2},\frac{13\pi}{2}, \frac{15\pi}{2}$

$\theta = \frac{\pi}{8}, \frac{3\pi}{8},\frac{5\pi}{8}, \frac{7\pi}{8},\frac{9\pi}{8}, \frac{11\pi}{8},\frac{13\pi}{8}, \frac{15\pi}{8}$

ebaines · Nov 1, 2010, 06:13 AM

Since $\sin \frac {\pi} 2 \pm 2 \pi n=1$ , where n = 0, 1, 2, 3, etc. then $\sin 4 \theta = 1$ for $\theta = \frac {\pi} 8 \pm \frac {\pi n} 2$ . Hence $\theta = \frac {\pi} 8 , \frac {5 \pi} 8, \frac {9 \pi} 8 ,$ , etc.

Similarly, $\sin \frac {3 \pi} 2 \pm 2 \pi n= -1$ , and so $\sin 4 \theta = -1$ for $\theta = \frac {3\pi} 8 \pm \frac {\pi } 2$ . Hence $\theta = \frac {3 \pi} 8 , \frac {7 \pi} 8, \frac {11 \pi} 8 ,$ , etc.

To draw your 8-pointed pedal, note that the radius for various angles are:

$ \begin {matrix} Angle & Radius \\ \\ 0 & 0 \\ \\ \frac {\pi} 8 & 5 \\ \\ \frac {\pi} 4 & 0 \\ \\ \frac {3 \pi} 8 &-5 \\ \\ \frac {\pi} 2 & 0 \\ \\ \frac {5 \pi} 8 & 5 \\ \\ \frac {3 \pi} 4 & 0 \\ \\ \frac {7 \pi} 8 & -5 \end{matrix}$