What are the components of the vector between the origin and the polar coordinate #(2, (-11pi)/6)#?

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Answer 1 · 2016-09-26T04:46:44

#(sqrt 3, 1)#

The components of the position vector to #r, theta) are (r cos theta, r sin theta)#

Here, they are

#(2 cos(-11/6pi), 2 sin (-11/6pi))#

#=(2 cos (11/6p)i, -2 sin (11/6pi))#,

using #cos(-a)=cos a and sin (-a)=-sin a#

#=(2 cos (2pi-pi/6), -2 sin (2pi-pi/6)#.

#=(2 cos (pi/6), 2 sin (pi/6))#,

using sine in #Q_4# is negative and cosine in #Q_4# is positive.

#=(sqrt 3, 1)#

You ought to note that, in effect, the directions

#theta = pi/6 and theta =-11/6pi=-(2pi-pi/6)#

are the same.

For positive #theta#, the sense of rotation is anticlockwise.

For negative #theta#, the sense of rotation is clockwise.