What is the Cartesian form of (r+3)^2 = sin^3theta-sec^2theta (r+3)2=sin3θsec2θ?

1 Answer
Feb 19, 2018

(r+3)^2=sin^3theta-sec^2theta(r+3)2=sin3θsec2θ when expressed in catesian form becomes

(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)(3+x2+y2)2=(yx2+y2)3(1+(yx)2)
which can be simplified as required

Explanation:

x=rcosthetax=rcosθ

y=rsinthetay=rsinθ

r=sqrt(x^2+y^2)r=x2+y2
r+3=sqrt(x^2+y^2)+3=3+sqrt(x^2+y^2)r+3=x2+y2+3=3+x2+y2
(r+3)^2=(3+sqrt(x^2+y^2))^2(r+3)2=(3+x2+y2)2

sin^3theta=(sintheta)^3sin3θ=(sinθ)3
sintheta=y/r=y/sqrt(x^2+y^2)sinθ=yr=yx2+y2
sin^3theta=(y/sqrt(x^2+y^2))^3sin3θ=(yx2+y2)3

sec^2theta=1+tan^2thetasec2θ=1+tan2θ
tan^2theta=(tantheta)^2tan2θ=(tanθ)2
theta=tan^-1(y/x)θ=tan1(yx)
tantheta=y/xtanθ=yx
tan^2theta=(y/x)^2tan2θ=(yx)2
sec^2theta=1+(y/x)^2sec2θ=1+(yx)2
Now, we have
(r+3)^2=(3+sqrt(x^2+y^2))^2(r+3)2=(3+x2+y2)2
sin^3theta=(y/sqrt(x^2+y^2))^3sin3θ=(yx2+y2)3
sec^2theta=1+(y/x)^2sec2θ=1+(yx)2

Thus,
(r+3)^2=sin^3theta-sec^2theta(r+3)2=sin3θsec2θ when expressed in catesian form becomes

(3+sqrt(x^2+y^2))^2=(y/sqrt(x^2+y^2))^3-(1+(y/x)^2)(3+x2+y2)2=(yx2+y2)3(1+(yx)2)
which can be simplified as required