How do you write $3 = r cos (θ - 60)$ $3=rcos(theta-60)$ in rectangular form?

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Answer 1 · 2016-11-29T11:13:34

$x + \sqrt{3} y = 6$ $x+sqrt3y=6$

The conversion formula is

$r (cos θ, sin θ) = (x, y)$ $r ( cos theta, sin theta ) = ( x, y )$

Here,

$3 = r (cos θ {cos 60}^{\oplus} sin θ {sin 60}^{0})$ $3 =r(cos theta cos 60^o+sin theta sin 60^0)$

$= \frac{1}{2} (r cos θ) + \frac{\sqrt{3}}{2} (r sin θ)$ $=1/2(r cos theta) +sqrt 3/2(r sin theta )$

$= \frac{1}{2} (x + \sqrt{3} y)$ $=1/2(x+sqrt3 y)$

How do you write 3=rcos(theta-60)3=rcos(θ−60)3=rcos(theta-60) in rectangular form?