The value - r -1 s sin2 -9- i cos2 -9 of isA- -1B- 1C- iD- - i

We have ∑_r=1^s(sin 2rπ/9+i cos 2rπ/9)=∑_r 1si (cos 2rπ/9 i sin 2rπ/9 ) i ∑_r=1^sa^r,when a= e^ (2ri/9) =ia (1 a^3)/(1 a)=(a a^9)/1 a=ia 1/1 a= i (∵ a^9=e^i ...

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Byju's Answer

Standard XII

Physics

Definite Integrals

The value ∑ r...

Question

The value $\sum_{r = 1}^{s} (s i n \frac{2 r π}{9} + i c o s \frac{2 r π}{9})$ of is

-1

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-i

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Solution

The correct option is D -i
we have
$\sum_{r = 1}^{s} (s i n \frac{2 r π}{9} + i c o s \frac{2 r π}{9}) = \sum_{r - 1} s i (c o s \frac{2 r π}{9} - i s i n \frac{2 r π}{9}) i \sum_{r = 1}^{s} a^{r}, w h e n a = e^{- (2 r i / 9)} = i a \frac{(1 - a^{3})}{(1 - a)} = \frac{(a - a^{9})}{1 - a} = i \frac{a - 1}{1 - a} = - i (∵ a^{9} = e^{i 2 π} = c o s 2 π - i s i n 2 π = 1)$

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The value ∑sr=1(sin2rπ9+i cos 2rπ9) of is

The correct option is D -iwe have ∑sr=1(sin2rπ9+i cos2rπ9)=∑r−1si(cos2rπ9−i sin 2rπ9)i ∑sr=1ar,when a=e−(2ri/9)=ia(1−a3)(1−a)=(a−a9)1−a=ia−11−a=−i(∵a9=ei2π=cos 2π−isin 2π=1)

The value $\sum_{r = 1}^{s} (s i n \frac{2 r π}{9} + i c o s \frac{2 r π}{9})$ of is