The points of intersections of $r = 2 \sin \theta$ and $r = 2 \cos \theta$ are

A) $\left( 1 , - \frac { \pi } { 3 } \right) , \left( 1 , \frac { 5 \pi } { 3 } \right)$
B) $\left( 1 , - \frac { \pi } { 3 } \right) , \left( 1 , \frac { 5 \pi } { 3 } \right) , ( 0,0 )$
C) $\left( \sqrt { 2 } , \frac { \pi } { 4 } \right) , ( 0,0 )$
D) $\left( 1 , \frac { 5 \pi } { 3 } \right) , ( 0,0 )$
E) $\left( \sqrt { 2 } , \frac { \pi } { 4 } \right)$

Question

The points of intersections of $r = 2 \sin \theta$ and $r = 2 \cos \theta$ are

A) $\left( 1 , - \frac { \pi } { 3 } \right) , \left( 1 , \frac { 5 \pi } { 3 } \right)$
B) $\left( 1 , - \frac { \pi } { 3 } \right) , \left( 1 , \frac { 5 \pi } { 3 } \right) , ( 0,0 )$
C) $\left( \sqrt { 2 } , \frac { \pi } { 4 } \right) , ( 0,0 )$
D) $\left( 1 , \frac { 5 \pi } { 3 } \right) , ( 0,0 )$
E) $\left( \sqrt { 2 } , \frac { \pi } { 4 } \right)$

Quizplus · Accepted Answer

To find the points of intersection, set $2\sin\theta = 2\cos\theta$. Dividing both sides by 2 and using the trigonometric identity $\tan\theta = \frac{\sin\theta}{\cos\theta}$, we get $\tan\theta = 1$, which has a solution at $\theta = \frac{\pi}{4}$. Substituting $\theta = \frac{\pi}{4}$ into either original equation gives $r = \sqrt{2}$. The origin $(0,0)$ is also a point of intersection since both equations are polar and intersect at the pole.

The Points of Intersections Of r=2sin⁡θr = 2 \sin \thetar=2sinθ And r=2cos⁡θr = 2 \cos \thetar=2cosθ

The Points of Intersections Of $r = 2 \sin \theta$ And $r = 2 \cos \theta$