# The value of (1 + sin2π/9 + i cos2π/9 / 1 + sin2π/9 − i cos2π/9)^3 is :

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The value of $\left({1+\sin{2\pi\over9}+i\cos{2\pi\over9}\over1+\sin{2\pi\over9}-i\cos{2\pi\over9}}\right)^3$ is :

(1) $-{1\over2}(\sqrt3-i)$

(2) ${1\over2}(\sqrt3-i)$

(3) ${1\over2}(1-i\sqrt3)$

(4) $-{1\over2}(1-i\sqrt3)$
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Ans: (1) $-{1\over2}(\sqrt3-i)$

Sol: $\left({1+\cos{5\pi\over18}+i\sin{5\pi\over18}\over1+\cos{5\pi\over18}-i\sin{5\pi\over18}}\right)^3$

$=\left({2\cos^2{5\pi\over36}+2i\sin{5\pi\over36}.\cos{5\pi\over36}\over2\cos^2{5\pi\over36}-2i\sin{5\pi\over36}.\cos{5\pi\over36}}\right)^3$

$=\left({\cos{5\pi\over36}+i\sin{5\pi\over36}\over\cos{5\pi\over36}-i\sin{5\pi\over36}}\right)^3$

$=\left(\cos{5\pi\over36}+i\sin{5\pi\over36}\right)^6$

$=\cos\left(6\times{5\pi\over36}\right)+i\sin\left(6\times{5\pi\over36}\right)$

$=\cos{5\pi\over6}+i\sin{5\pi\over6}$

$=-{\sqrt3\over2}+i{1\over2}$

$=-{1\over2}(\sqrt3-i)$
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Most helping solution in Trigonometry.
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