If w is a root of the equation x 2- x -1-0- The expression An-r-1nr-wr-w2 and Bn-r-1nr-wr-w2- then the value of sin-An-Bn -n-

X^2 + x+ 1 nbsp; = 0 x = 1 ± i √(3)2 w = 1 nbsp; + i √(3)2 which is a cube root of unity. ∴ 1 + w+ w^2 = 0 and w^3 = 1 A_n = ∑_r = 1^n(r w)(r w^2) ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

If w is a roo...

Question

If $w$ is a root of the equation $x^{2} + x + 1 = 0$ . The expression
$A_{n} = n \sum r = 1 (r - w) (r - w^{2})$ and $B_{n} = n \sum r = 1 (r + w) (r + w^{2})$ , then the value of $sin [(A_{n} - B_{n}) \cdot \frac{π}{n}]$ is

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Solution

$x^{2} + x + 1 = 0 x = \frac{- 1 \pm i \sqrt{3}}{2}$
$w = \frac{- 1 + i \sqrt{3}}{2}$ which is a cube root of unity.
$∴ 1 + w + w^{2} = 0$ and $w^{3} = 1$

$A_{n} = n \sum r = 1 (r - w) (r - w^{2}) = n \sum r = 1 (r^{2} - (w + w^{2}) r + w^{3}) = n \sum r = 1 (r^{2} + r + 1)$

$B_{n} = n \sum r = 1 (r + w) (r + w^{2}) = n \sum r = 1 (r^{2} + (w + w^{2}) r + w^{3}) = n \sum r = 1 (r^{2} - r + 1)$

$A_{n} - B_{n} = n \sum r = 1 2 r = (n) (n + 1)$
Hence,
$sin [(A_{n} - B_{n}) \cdot \frac{π}{n}] = sin (n + 1) π = 0$

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If w is a root of the equation x2+x+1=0. The expression An=n∑r=1(r−w)(r−w2) and Bn=n∑r=1(r+w)(r+w2), then the value of sin[(An−Bn)⋅πn] is

x2+x+1=0x=−1±i√32 w=−1+i√32 which is a cube root of unity. ∴1+w+w2=0 and w3=1 An=n∑r=1(r−w)(r−w2) =n∑r=1(r2−(w+w2)r+w3) =n∑r=1(r2+r+1) Bn=n∑r=1(r+w)(r+w2) =n∑r=1(r2+(w+w2)r+w3) =n∑r=1(r2−r+1) An−Bn=n∑r=12r=(n)(n+1) Hence, sin[(An−Bn)⋅πn]=sin(n+1)π=0

If $w$ is a root of the equation $x^{2} + x + 1 = 0$ . The expression
$A_{n} = n \sum r = 1 (r - w) (r - w^{2})$ and $B_{n} = n \sum r = 1 (r + w) (r + w^{2})$ , then the value of $sin [(A_{n} - B_{n}) \cdot \frac{π}{n}]$ is