If - is a cube root of unity- then sin-900-r-110r-r-2-A- -1B- 0C- -3-2

∑_r=1^10(r ω)(r ω^2)=∑_r=1^10(r^2 r( 1)+1) =∑_r=1^10r^2+r+1 =10(10+1)(2× 10+1)6+10(10+1)2+10 =10 · 11 · 216+10 · 112+10= 450 ∴sin( 450 ·π900)=sinπ nbsp; 2=1 ...

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Standard XII

Mathematics

Cube Root of a Complex Number

If ω is a cub...

Question

If $ω$ is a cube root of unity, then $sin \frac{π}{900} {10 \sum r = 1 (r - ω) (r - ω^{2})} =$

- 1

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\frac{\sqrt{3}}{2}

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Solution

The correct option is C $1$
$10 \sum r = 1 (r - ω) (r - ω^{2}) = 10 \sum r = 1 (r^{2} - r (- 1) + 1)$
$= 10 \sum r = 1 r^{2} + r + 1$
$= \frac{10 (10 + 1) (2 \times 10 + 1)}{6} + \frac{10 (10 + 1)}{2} + 10$
$= \frac{10 \cdot 11 \cdot 21}{6} + \frac{10 \cdot 11}{2} + 10 = 450$
$∴ sin (450 \cdot \frac{π}{900}) = sin \frac{π}{2} = 1$

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If ω is a cube root of unity, then sinπ900{10∑r=1(r−ω)(r−ω2)}=

The correct option is C 110∑r=1(r−ω)(r−ω2)=10∑r=1(r2−r(−1)+1) =10∑r=1r2+r+1 =10(10+1)(2×10+1)6+10(10+1)2+10 =10⋅11⋅216+10⋅112+10=450 ∴sin(450⋅π900)=sinπ2=1

If $ω$ is a cube root of unity, then $sin \frac{π}{900} {10 \sum r = 1 (r - ω) (r - ω^{2})} =$