The trigonometric form of z-1-i cot83 where i-1 is

Z=(1 i cot8)^3 =cosec^38(sin 8 i cos 8)^3 =cosec^38 ( cos ( π/2 8 ) i sin ( π/2 8 ) )^3 =cosec^38 (cos ( 3π/2 24 ) i sin ( 3π/2 24 ) ) =cosec^38.e^ i ...

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Byju's Answer

Standard XII

Mathematics

Polar Representation of a Complex Number

The trigonome...

Question

The trigonometric form of $z = (1 - i c o t 8)^{3}$ (where $i = \sqrt{- 1}$ ) is

$c o s e c^{3} 8. e^{i (24 - \frac{3 π}{2})}$

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$c o s e c^{3} 8. e^{- i (24 - \frac{3 π}{2})}$

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$c o s e c^{3} 8. e^{i (36 - \frac{π}{2})}$

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$c o s e c^{2} 8. e^{- (24 i \frac{+ π}{2})}$

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Solution

The correct option is A
$c o s e c^{3} 8. e^{i (24 - \frac{3 π}{2})}$

$z = (1 - i c o t 8)^{3} = c o s e c^{3} 8 (sin 8 - i cos 8)^{3} = c o s e c^{3} 8 {(cos (\frac{π}{2} - 8) - i sin (\frac{π}{2} - 8))}^{3} = c o s e c^{3} 8 (cos (\frac{3 π}{2} - 24) - i sin (\frac{3 π}{2} - 24)) = c o s e c^{3} 8. e^{- i (\frac{3 π}{2} - 24)} = c o s e c^{3} 8. e^{i (24 - \frac{3 π}{2})}$

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The trigonometric form of z=(1−i cot8)3 (where i=√−1) is

The correct option is A cosec38.ei(24−3π2) z=(1−i cot8)3=cosec38(sin8−i cos8)3=cosec38(cos(π2−8)−i sin(π2−8))3=cosec38(cos(3π2−24)−i sin(3π2−24))=cosec38.e−i(3π2−24)=cosec38.ei(24−3π2)

The trigonometric form of $z = (1 - i c o t 8)^{3}$ (where $i = \sqrt{- 1}$ ) is