What is the Cartesian form of $(71, \frac{13 π}{12})$ $(71,(13pi) /12)$ ?

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Answer 1 · 2016-01-22T08:54:44

$(- \frac{71 \cdot \sqrt{2 + \sqrt{3}}}{2}, - \frac{71 \cdot \sqrt{2 - \sqrt{3}}}{2})$ $(-(71*sqrt(2+sqrt(3)))/2, -(71*sqrt(2-sqrt(3)))/2)$

$x = r \cdot cos θ$ $x=r*cos theta$ and $y = r \cdot sin θ$ $y=r*sin theta$

Note: that $\frac{π}{12} = \frac{\frac{π}{6}}{2}$ $pi/12=(pi/6)/2$

Use half angle formulas
$sin (\frac{A}{2}) = \sqrt{\frac{1 - cos A}{2}}$ $sin (A/2)=sqrt((1-cos A)/2)$

and $cos (\frac{A}{2}) = \sqrt{\frac{1 + cos A}{2}}$ $cos (A/2)=sqrt((1+cos A)/2)$

$x = 71 \cdot cos (\frac{13 π}{12})$ $x=71*cos ((13pi)/12)$

$x = 71 \cdot (- \frac{\sqrt{2 + \sqrt{3}}}{2})$ $x=71*(-sqrt(2+sqrt(3))/2)$

$x = - \frac{71 \cdot \sqrt{2 + \sqrt{3}}}{2}$ $x=-(71*sqrt(2+sqrt(3)))/2$

$y = 71 \cdot sin (\frac{13 π}{12})$ $y=71*sin ((13pi)/12)$

$y = 71 \cdot (- \frac{\sqrt{2 - \sqrt{3}}}{2})$ $y=71*(-sqrt(2-sqrt(3))/2)$

$y = - \frac{71 \cdot \sqrt{2 - \sqrt{3}}}{2}$ $y= -(71*sqrt(2-sqrt(3)))/2$

What is the Cartesian form of (71,13π12)(71,(13pi) /12)?