What are the components of the vector between the origin and the polar coordinate $(- 4, \frac{- 7 π}{4})$ $(-4, (-7pi)/4)$ ?

Question

1455 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T01:40:33

$< - 2 \sqrt{2}, - 2 \sqrt{2} >$ $<-2sqrt2, -2sqrt 2>$ .-

In both cartesian $(x, y) and$ $(x, y) and$ polar $(r, θ)$ $(r, theta)$ forms, the

components of the position vector $O P$ $OP$ , from the origin to the point

P(x,y) are $< x, y > = < r (cos θ, sin θ) >$ $< x, y > = < r (cos theta, sin theta)>$ , repectively

Here, $(r, θ) = (- 4, - \frac{7 π}{4})$ $(r, theta)=(-4, -(7pi)/4)$ . and so, the components are

$< - 4 cos (- 7 \frac{π}{4}), - 4 sin (- 7 \frac{π}{4}) >$ $<-4 cos(-7pi/4), -4 sin(-7pi/4)>$ , using cos (-a) = cos a and sin (-a)

=-sin a

$= < - 4 cos (7 \frac{π}{4}), 4 sin (7 \frac{π}{4}) >$ $= <-4cos(7pi/4), 4 sin (7pi/4)>$

$= < - 4 cos (2 π - \frac{π}{4}), 4 sin (2 π - \frac{π}{4}) >$ $= <-4 cos (2pi-pi/4), 4 sin (2pi-pi/4) >$

$= < - 4 cos (\frac{π}{4}) - 4 sin (\frac{π}{4}) >$ $= <-4 cos (pi/4)- 4 sin (pi/4) >$ ,

using $cos (2 π - a) = cos a and sin (2 π - a) = - sin a$ $cos (2pi-a)=cos a and sin (2pi-a)=-sina$ .

$= < - \frac{4}{\sqrt{2}}, - \frac{4}{\sqrt{2}} >$ $= <-4/sqrt2, -4/sqrt2 >$

$= < - 2 \sqrt{2}, - 2 \sqrt{2} >$ $= <-2sqrt2, -2sqrt 2>$ .-

What are the components of the vector between the origin and the polar coordinate (-4, (-7pi)/4)(−4,−7π4)(-4, (-7pi)/4)?