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:

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

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