The principle amplitude of $(sin 40^{o} + i cos 40^{o})^{5}$ is

70^{o}

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- 1100^{o}

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70^{110}

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70^{- 70}

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Solution

The correct option is D $70^{o}$
$40 ° = 2 \frac{π}{9}, 50 ° = \frac{5 π}{18}$

So, $(sin 40 ° + i cos 40 °) = cos 50 ° + i sin 50 °$

Let $T = {(sin 40 ° + i cos 40 °)}^{5} = {(cos 50 ° + i sin 50 °)}^{5}$

$T = {(e^{i 50 °})}^{5} = ⎛ ⎜ ⎝ e^{i \frac{5 π}{18} \times 5} ⎞ ⎟ ⎠ = e^{i \frac{25 π}{18}}$

So, $25 \frac{π}{18} = 250 °$

Thus, principal argument $= - 180 ° + 250 ° = 70 °$

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The principle amplitude of (sin40o+icos40o)5 is

The correct option is D 70o40°=2π9,50°=5π18So, (sin40°+icos40°)=cos50°+isin50°Let T=(sin40°+icos40°)5=(cos50°+isin50°)5T=(ei50°)5=⎛⎜⎝ei5π18×5⎞⎟⎠=ei25π18So, 25π18=250°Thus, principal argument=−180°+250°=70°

The principle amplitude of $(sin 40^{o} + i cos 40^{o})^{5}$ is