Respuesta :
[tex]x^2+y^2=r^2~\hfill x=rcos(\theta )~\hfill y=rsin(\theta ) \\\\[-0.35em] ~\dotfill\\\\ r=3sin(A)\implies \stackrel{\textit{multiplying both sides by "r"}}{r^2~~ = ~~3~rsin(A)}\implies x^2+y^2=3y[/tex]
r = 3sinA is written in rectangular form as[tex]x^2+y^2 = 3y.[/tex]
What is De Moivre's theorem for exponentiation of complex numbers?
Any complex number z = a + ib can be written in polar form as:
[tex]z = r(\cos(\theta) + i\sin(\theta))[/tex]
Complex numbers are those numbers that contain the imaginary and the real part.
Raising this to nth power (n being an integer), we get:
[tex]z ^n = r^n (\cos(n\theta) + i\sin(n\theta))[/tex]
We need to find r = 3sinA is written in rectangular form as[tex]x^2+y^2[/tex].
We know that
x = r cosA
y = r sin A
We have been given ;
r = 3sinA
Multiply by r on both sides;
[tex]r^2 = 3r sin A\\\\r^2 = 3 y[/tex]
therefore,
[tex]x^2+y^2 = 3y[/tex]
Hence, r = 3sinA is written in rectangular form as [tex]x^2+y^2 = 3y[/tex].
Learn more about cube root of complex numbers here:
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