Can you find [ (2(cos80˚+isin80˚))^3 ] ? then express the answer in rectangular form.

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Can you find [ (2(cos80˚+isin80˚))^3 ] ? then express the answer in rectangular form.

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Answer 1 · 2018-07-26T18:31:33

$(2 {(({cos 80}^{0} + i {sin 80}^{0}))}^{3} = - 4 - 4 i \sqrt{3}$ $(2((cos80^0+isin80^0))^3=-4-4isqrt3$

Explanation:

$(2 {(({cos 80}^{0} + i {sin 80}^{0}))}^{3} = 2^{3} {({cos 80}^{0} + i {sin 80}^{0})}^{3}$ $(2((cos80^0+isin80^0))^3=2^3(cos80^0+isin80^0)^3$
By De-Moivre's theorem;
${(cos A + i sin A)}^{n} = cos n A + i sin n A$ $(cosA+isinA)^n=cosnA+isinnA$
$= 8 (cos (3 \times 80^{0}) + i sin (3 \times 80^{0}))$ $=8(cos(3xx80^0)+isin(3xx80^0))$
$= 8 ({cos 240}^{0} + i {sin 240}^{0})$ $=8(cos240^0+isin240^0)$
240=180+60
${cos 240}^{0} = cos (180^{0} + 60^{0})$ $cos240^0=cos(180^0+60^0)$
$= - {cos 60}^{0} = - \frac{1}{2}$ $=-cos60^0=-1/2$
${sin 240}^{0} = sin (180^{0} + 60^{0})$ $sin240^0=sin(180^0+60^0)$
$= - {sin 60}^{0} = - \frac{\sqrt{3}}{2}$ $=-sin60^0=-sqrt3/2$
Thus,
$8 ({cos 240}^{0} + i {sin 240}^{0}) = 8 \times (- \frac{1}{2} + i \frac{- \sqrt{3}}{2})$ $8(cos240^0+isin240^0)=8xx(-1/2+i(-sqrt3)/2)$
$= - 4 - 4 i \sqrt{3}$ $=-4-4isqrt3$

$(2 {(({cos 80}^{0} + i {sin 80}^{0}))}^{3} = - 4 - 4 i \sqrt{3}$ $(2((cos80^0+isin80^0))^3=-4-4isqrt3$

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Can you find [ (2(cos80˚+isin80˚))^3 ] ? then express the answer in rectangular form.

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