Question

find the distance between the points A cos theta a sin theta and minus a sin theta minus cos theta

Answer 1

OK I send you later okay

Answer 2

Answer:

 $AB= \sqrt{2}a$

Step-by-step explanation:

A=$(x_1,y_1)=(a cos \theta , a sin \theta)$

B = $(x_2,y_2)=(a sin \theta , -a cos \theta)$

Distance formula : $d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the values

$AB= \sqrt{(a sin \theta-a cos \theta)^2+(-a cos \theta-a sin \theta)^2}$

 $AB= \sqrt{a^2(sin^2 \theta+cos^2 \theta-2 sin\theta cos \theta )+a^2( cos^2 \theta+ sin^2\theta+2 sin\theta cos \theta)}$

 $AB= \sqrt{a^2(1-2 sin\theta cos \theta )+a^2( 1+2 sin\theta cos \theta)}$

 $AB= \sqrt{2}a$