How do you prove that $2 cos (θ + \frac{π}{3}) = cos θ - \sqrt{3} sin θ$ $2cos(theta+pi/3) = costheta-sqrt3sintheta$ ?

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Answer 1 · 2018-05-26T03:16:58

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$2 cos (θ + \frac{π}{3}) = cos θ - \sqrt{3} sin θ$ $2cos(theta+pi/3)=costheta-sqrt3sintheta$

LHS
$2 cos (θ + \frac{π}{3})$ $2cos(theta+pi/3)$

Recall: $cos (a - b) = cos a cos b - sin a sin b$ $cos(a-b)=cosacosb-sinasinb$

= $2 (cos θ cos (\frac{π}{3}) - sin θ sin (\frac{π}{3}))$ $2(costhetacos(pi/3)-sinthetasin(pi/3))$

= $2 (cos θ \times \frac{1}{2} - sin θ \times \frac{\sqrt{3}}{2})$ $2(costhetatimes1/2-sinthetatimessqrt3/2)$

= $2 \times \frac{1}{2} \times cos θ - \frac{1}{2} \times \frac{\sqrt{3}}{2} \times sin θ$ $2times1/2timescostheta-1/2timessqrt3/2timessintheta$

= $cos θ - \sqrt{3} sin θ$ $costheta-sqrt3sintheta$

= RHS

How do you prove that 2cos(θ+π3)=cosθ−√3sinθ2cos(theta+pi/3) = costheta-sqrt3sintheta?