If S-cos 2-n-cos 22 -n- - -cos 2n-1 -n then S is equal to

The correct option is B S = 1/2[1 + cos2π/n + 1 + cos4π/n + 1 + nbsp;cos6π/n + ...+ ..... + 1 + nbsp;cos2(n 1)π/n] nbsp; nbsp;= nbsp;1/2 [ (n 1) + ...

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Byju's Answer

Standard XII

Mathematics

Converting to Perfect Square

If S=cos 2π/n...

Question

If S = $c o s^{2} \frac{π}{n} + c o s^{2} \frac{2 π}{n} + . . . . . . . . + c o s^{2} \frac{(n - 1) π}{n}$ then S is equal to

$\frac{n(n-2)}{2}$

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$\frac{(n-1)}{2}$

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$\frac{(n-2)}{2}$

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n/2

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Solution

The correct option is C
$\frac{(n-2)}{2}$

S = $\frac{1}{2} [1 + c o s \frac{2 π}{n} + 1 + c o s \frac{4 π}{n} + 1 + c o s \frac{6 π}{n} + . . . + . . . . . + 1 + c o s 2 (n - 1) \frac{π}{n}]$

= $\frac{1}{2} [(n - 1) + \sum_{λ = 1}^{n - 1} c o s \frac{2 k π}{n}]$

= $\frac{1}{2} [n - 1 - 1] = \frac{1}{2} (n - 2)$

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If S = cos2πn+cos22πn+........+cos2(n−1)πn then S is equal to

The correct option is C S = 12[1+cos2πn+1+cos4πn+1+cos6πn+...+.....+1+cos2(n−1)πn] = 12[(n−1)+∑n−1λ=1cos2kπn] = 12[n−1−1]=12(n−2)

If S = $c o s^{2} \frac{π}{n} + c o s^{2} \frac{2 π}{n} + . . . . . . . . + c o s^{2} \frac{(n - 1) π}{n}$ then S is equal to