How do you graph $r = 2 csc (θ + 60)$ $r=2csc(theta+60)$ ?

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Answer 1 · 2016-12-09T13:36:19

A straight line that makes intercepts $\frac{4}{\sqrt{3}} and 4$ $4/sqrt3 and 4$ on $θ = 0 and θ = 90^{o}$ $theta=0 and theta=90^o$

This is the polar equation of the straight line $\frac{x}{\frac{4}{\sqrt{3}}} + \frac{y}{4} = 1$ $x/(4/sqrt3)+y/4=1$ that

makes intercepts $\frac{4}{\sqrt{3}} and 4$ $4/sqrt3 and 4$ on^theta=0 and theta=pi/2..

Using cos theta =x/r and sin theta = y/r,

rsin(theta+60^o)#

$= r (sin θ {cos 60}^{o} + cos θ {sin 60}^{0})$ $=r(sintheta cos 60^o +cos theta sin 60^0)$

$= \frac{y}{2} + \frac{\sqrt{3}}{2} x = 2$ $=y/2+sqrt3/2x=2$ and this, in the intercept form, is

$\frac{x}{\frac{4}{\sqrt{3}}} + \frac{y}{4} = 1$ $x/(4/sqrt3)+y/4=1$

graph{y+sqrt3 x=4 [-10, 10, -5, 5]}

How do you graph r=2csc(θ+60)r=2csc(theta+60)?