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Polar to Rectangular Equations
- Thread starter dulaney11
- Start date
Note that \(\displaystyle y=rsin{\theta}, \;\ x=rcos{\theta}\)
\(\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}=sin{\theta}\) and so on.
You can express \(\displaystyle sin(3{\theta})=4sin{\theta}cos^{2}{\theta}-sin{\theta}\)
and go from there.
\(\displaystyle \frac{y}{\sqrt{x^{2}+y^{2}}}=sin{\theta}\) and so on.
You can express \(\displaystyle sin(3{\theta})=4sin{\theta}cos^{2}{\theta}-sin{\theta}\)
and go from there.
I had some misplaced parentheses. I fixed it.
It comes from the addition formula for sine.
\(\displaystyle sin(3x)=sin(x+2x)=sin(x)cos(2x)+cos(x)sin(2x)\)
\(\displaystyle =sin(x)(cos^{2}(x)-sin^{2}(x))+2sin(x)cos^{2}(x)\)
And it whittles down to
\(\displaystyle 4sin(x)cos^{2}(x)-sin(x)\)
\(\displaystyle \sqrt{x^{2}+y^{2}}=2\left[4\left(\frac{y}{\sqrt{x^{2}+y^{2}}}\right)\left(\frac{x^{2}}{x^{2}+y^{2}}\right)-\frac{y}{\sqrt{x^{2}+y^{2}}}\right]\)
Simplify
It comes from the addition formula for sine.
\(\displaystyle sin(3x)=sin(x+2x)=sin(x)cos(2x)+cos(x)sin(2x)\)
\(\displaystyle =sin(x)(cos^{2}(x)-sin^{2}(x))+2sin(x)cos^{2}(x)\)
And it whittles down to
\(\displaystyle 4sin(x)cos^{2}(x)-sin(x)\)
\(\displaystyle \sqrt{x^{2}+y^{2}}=2\left[4\left(\frac{y}{\sqrt{x^{2}+y^{2}}}\right)\left(\frac{x^{2}}{x^{2}+y^{2}}\right)-\frac{y}{\sqrt{x^{2}+y^{2}}}\right]\)
Simplify
D
Deleted member 4993
Guest
dulaney11 said:Okay, back to the whole question, I can get
r= 2 [sinx(cos^2x-sin^x)] + 2 sinxcos^2x <<< This is so off - it is not even wrong. Study Galactus's response carefully. He almost solved it for you.
distribute- r= 2sinx2cos^x-2sin^2x + 2sinxcos^x
So therefore, from that I don't understand how to simplify that into rectangular form
D
Deleted member 4993
Guest
dulaney11 said:Thanks, but I am not quite sure how you would have gotten that? <<< By studying trigonometry in high school