The equation r cos left(theta-frac{pi}{3}righ | vedclass

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Question

(D) Given the polar equation: $r \cos \left(	heta-\frac{\pi}{3}ight)=2$ Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B

Vedclass · Accepted Answer

a straight line

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(D) Given the polar equation: $r \cos \left(	heta-\frac{\pi}{3}ight)=2$ Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$: $r \left( \cos 	heta \cos \frac{\pi}{3} + \sin 	heta \sin \frac{\pi}{3} ight) = 2$ Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$: $r \left( \cos 	heta \cdot \frac{1}{2} + \sin 	heta \cdot \frac{\sqrt{3}}{2} ight) = 2$ Multiplying by $2$: $r \cos 	heta + \sqrt{3} r \sin 	heta = 4$ Using the conversion formulas $x = r \cos 	heta$ and $y = r \sin 	heta$: $x + \sqrt{3} y = 4$ This is a linear equation in $x$ and $y$, which represents a straight line.

The equation $r \cos \left(\theta-\frac{\pi}{3}\right)=2$ represents

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