Polar Coordinates to Rectangular Coordinates

[nycmath]

See attachment. Enjoy.

[Eigenvalue]

x=4*cos(11π/6)=4*(1/2)=2
x=4*sin(11π/6)=4(-√3/2)=-2√3
(2, -2√3)

π/8=(1/2)*(π/4)
cos(π/8)=[tex]\pm \sqrt{ \frac{1+cos(π/4}{2} }[/tex]

Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

sin(π/8)=[tex]\pm \sqrt{ \frac{1-cos(π/4}{2} }[/tex]


Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

x=-1*cos(π/8)=-[tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

y=-1*sin(theta)=- [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

[nycmath]

x=4*cos(11π/6)=4*(1/2)=2
x=4*sin(11π/6)=4(-√3/2)=-2√3
(2, -2√3)

π/8=(1/2)*(π/4)
cos(π/8)=[tex]\pm \sqrt{ \frac{1+cos(π/4}{2} }[/tex]

Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

sin(π/8)=[tex]\pm \sqrt{ \frac{1-cos(π/4}{2} }[/tex]


Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

x=-1*cos(π/8)=-[tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

y=-1*sin(theta)=- [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

What formula did you use in both examples?

[Eigenvalue]

x=4*cos(11π/6)=4*(1/2)=2
x=4*sin(11π/6)=4(-√3/2)=-2√3
(2, -2√3)

π/8=(1/2)*(π/4)
cos(π/8)=[tex]\pm \sqrt{ \frac{1+cos(π/4}{2} }[/tex]

Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

sin(π/8)=[tex]\pm \sqrt{ \frac{1-cos(π/4}{2} }[/tex]


Since π/8 is in the first quadrant, it equals to [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

x=-1*cos(π/8)=-[tex]\frac{ \sqrt{2+ \sqrt{2} } }{2}[/tex]

y=-1*sin(theta)=- [tex]\frac{ \sqrt{2-\sqrt{2} } }{2}[/tex]

What formula did you use in both examples?

Half angle formulas for sine and cosine