How do you calculate $3 ({cos 14}^{\circ} + i {sin 14}^{\circ}) \cdot 2 ({cos 121}^{\circ} + i {sin 121}^{\circ})$ $3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ)$ ?

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How do you calculate $3 ({cos 14}^{\circ} + i {sin 14}^{\circ}) \cdot 2 ({cos 121}^{\circ} + i {sin 121}^{\circ})$ $3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ)$ ?

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Answer 1 · 2017-04-06T16:23:08

Multiplication this type is defined as:

$r_{1} (cos (θ_{1}) + i sin (θ_{1}) \cdot 2 (cos (θ_{2}) + i sin (θ_{2}) = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$ $r_1(cos(theta_1)+isin(theta_1)*2(cos(theta_2)+isin(theta_2)=r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))$

Explanation:

In this case, $r_{1} = 3, r_{2} = 2, θ_{1} = 14^{\circ}, and θ_{2} = 121^{\circ}$ $r_1 = 3, r_2 = 2, theta_1 = 14^@", and "theta_2 =121^@$ :

$3 ({cos 14}^{\circ} + i {sin 14}^{\circ}) \cdot 2 ({cos 121}^{\circ} + i {sin 121}^{\circ}) = 6 (cos (135^{\circ}) + i sin (135^{\circ}))$ $3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ) = 6(cos(135^@)+isin(135^@))$

This simplifies to:

$- 3 \sqrt{2} + 3 \sqrt{2} i$ $-3sqrt2+3sqrt2i$

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How do you calculate $3 ({cos 14}^{\circ} + i {sin 14}^{\circ}) \cdot 2 ({cos 121}^{\circ} + i {sin 121}^{\circ})$ $3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ)$ ?

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Explanation:

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How do you calculate $3 ({cos 14}^{\circ} + i {sin 14}^{\circ}) \cdot 2 ({cos 121}^{\circ} + i {sin 121}^{\circ})$ $3(cos14^circ+isin14^circ)*2(cos121^circ+isin121^circ)$ ?