How do you write the equation #6=rcos(theta-120^circ)# in rectangular form?

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Answer 1 · 2016-11-29T22:41:51

Please see the explanation.

Use the identity #cos(A - B) = cos(A)cos(B) + sin(A)sin(B)#

#cos(theta - 120^@) = cos(theta)cos(120^@) + sin(theta)sin(120^@)#

#cos(theta - 120^@) = -1/2cos(theta) + sqrt(3)/2sin(theta)#

Substitute into the given equation:

#6 = r(-1/2cos(theta) + sqrt(3)/2sin(theta))#

Distribute r through the ()s:

#6 = -1/2rcos(theta) + sqrt(3)/2rsin(theta)#

Substitute x for #rcos(theta)# and y for #rsin(theta)#

#6 = -1/2x + sqrt(3)/2y#

Multiply both sides by -2:

#x - (sqrt(3))y = -12#