What are the components of the vector between the origin and the polar coordinate #(5, pi/12)#?

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Answer 1 · 2017-12-30T22:52:00

#v_x = 5 sqrt(2+sqrt(3))/2#; #" "v_y= 5 sqrt(2-sqrt(3))/2#

Given: polar vector from #(0,0)# to #(5, pi/12)#

This vector is in the first quadrant.

A polar vector: #(r, theta) = (5, pi/12)#

#x#-component #= r cos theta = 5 cos (pi/12)#

#y#-component #= r sin theta = 5 sin (pi/12)#

Using Half-angle formulas you can find the exact value of the angle #pi/12 = (pi/6)/2# :

#cos (a/2) = +- sqrt((1+cos a)/2); " "sin (a/2) = +- sqrt((1-cos a)/2)#

#cos ((pi/6)/2) = sqrt((1+cos (pi/6))/2)#; #" "sin ((pi/6)/2) = sqrt((1-cos (pi/6))/2)#

#cos (pi/6) = cos 30^o = sqrt(3)/2#

#cos(pi/12) = sqrt((1+sqrt(3)/2)/2)#; #" "sin(pi/12) = sqrt((1-sqrt(3)/2)/2)#

#cos(pi/12) = sqrt(((2+sqrt(3))/2)*1/2) = sqrt(2+sqrt(3))/2#

#sin(pi/12) = sqrt(((2-sqrt(3))/2)*1/2) = sqrt(2-sqrt(3))/2#

#x#-component #= r cos theta = 5 sqrt(2+sqrt(3))/2#

#y#-component #=r sin theta = 5 sqrt(2-sqrt(3))/2#