Question #49b10

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Question #49b10

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Answer 1 · 2018-01-20T16:33:38

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

Explanation:

The general form is $a \cdot cos (θ) - b \cdot sin (θ) = R \cdot cos (θ + α)$ $a*cos(theta)-b*sin(theta) = R*cos(theta+alpha)$ , where $R = \sqrt{a^{2} + b^{2}}$ $R = sqrt(a^2+b^2)$ and $α = {tan}^{- 1} (\frac{b}{a})$ $alpha=tan^-1(b/a)$ .

First calculate $R = \sqrt{a^{2} + b^{2}} = \sqrt{{(\sqrt{3})}^{2} + {(3)}^{2}} = \sqrt{12} = 2 \sqrt{3}$ $R = sqrt(a^2+b^2)= sqrt((sqrt(3))^2+(3)^2)=sqrt(12)=2sqrt(3)$ .

Next calculate $α = {tan}^{- 1} (\frac{b}{a}) = {tan}^{- 1} (\frac{3}{\sqrt{3}}) = {tan}^{- 1} (\sqrt{3}) = \frac{π}{3}$ $alpha=tan^-1(b/a) = tan^-1(3/sqrt(3))=tan^(-1)(sqrt(3)) = pi/3$ .

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

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Question #49b10

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