Sec theta = 2 to rectangular coordinates??

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Answer 1 · 2018-02-26T15:23:06

See below.

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Identity.

$sec x = \frac{1}{cos x}$ $color(red)bb(secx=1/cosx)$

$\frac{1}{cos (θ)} = 2$ $1/cos(theta)=2$

$cos (θ) = \frac{1}{2}$ $cos(theta)=1/2$

$θ = arccos (cos (θ)) = arccos (\frac{1}{2}) \Rightarrow θ = \frac{π}{3}, \frac{5 π}{3}$ $theta=arccos(cos(theta))=arccos(1/2)=>theta=pi/3, (5pi)/3$

For:

$0 \leq θ \leq 2 π$ $0<=theta<=2pi$

From the above diagram, we can see that:

$x = r cos (θ)$ $x=rcos(theta)$

$y = r sin (θ)$ $y=rsin(theta)$

So coordinates will be:

$(r cos (\frac{π}{3}), r sin (\frac{π}{3}))$ $(rcos(pi/3),rsin(pi/3))$

$(r cos (\frac{5 π}{3}), r sin (\frac{5 π}{3}))$ $(rcos((5pi)/3),rsin((5pi)/3))$

Without knowing the radius we can't go any further than this.

If this is on a unit circle then the radius is $1$ $1$ and:

$cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3)=1/2$ , $sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3)=sqrt(3)/2$

$cos (\frac{5 π}{3}) = \frac{1}{2}$ $cos((5pi)/3)=1/2$ , $sin (\frac{5 π}{3}) = - \frac{\sqrt{3}}{2}$ $sin((5pi)/3)=-sqrt(3)/2$

Coordinates:

$(\frac{1}{2}, \frac{\sqrt{3}}{2})$ $(1/2,sqrt(3)/2)$ , $(\frac{1}{2}, - \frac{\sqrt{3}}{2})$ $(1/2 , -sqrt(3)/2)$