How do you multiply $(5 - i) (6 - 4 i)$ $(5-i)(6-4i)$ in trigonometric form?

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How do you multiply $(5 - i) (6 - 4 i)$ $(5-i)(6-4i)$ in trigonometric form?

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Answer 1 · 2016-04-10T07:05:24

$C = \sqrt{26} \cdot \sqrt{52} ∠ (349^{0} + 326^{0})$ $C= sqrt(26)*sqrt(52)/_(349^0+326^0)$
$C \approx 36.8 ∠ 315^{0}$ $C~~36.8/_315^0$

Explanation:

Given: $C_{1} = (5 - i), C_{2} = (6 - 4 i)$ $C_1= (5-i), C_2=(6-4i)$

Required: The product of $C_{1} \cdot C_{2}$ $C_1*C_2$ in trigonometric form

Solution Strategy:
1) Convert the Phasors (another name for complex numbers) to their polar or trig form. That is given by:
$C_{1} = | C_{1} | (cos θ + i sin θ)$ $C_1 = |C_1| (cos theta +isintheta)$
where: $| C_{1} | = \sqrt{5^{2} + 1^{2}} = \sqrt{26}; θ = {tan}^{- 1} (- \frac{1}{5}) = 349$ $|C_1|= sqrt(5^2+1^2)=sqrt(26); theta= tan^-1 (-1/5)=349$

$C_{2} = | C_{2} | (cos α + i sin α)$ $C_2 = |C_2| (cos alpha+isinalpha)$
$| C_{2} | = \sqrt{6^{2} + 4^{2}} = \sqrt{52}; θ = {tan}^{- 1} (- \frac{4}{6}) = 326$ $|C_2|= sqrt(6^2+4^2)=sqrt(52); theta= tan^-1 (-4/6)=326$

2) Product: $C_{1} \cdot C_{2} = \sqrt{26} ∠ 349^{o} \cdot \sqrt{52} ∠ 326^{o}$ $C_1*C_2= sqrt(26)/_349^o*sqrt(52)/_326^o$
$= \sqrt{26} \cdot \sqrt{52} ∠ (349^{0} + 326^{0})$ $= sqrt(26)*sqrt(52)/_(349^0+326^0)$

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How do you multiply $(5 - i) (6 - 4 i)$ $(5-i)(6-4i)$ in trigonometric form?

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