The value of - n -110sin2 n -11-cos2 n -11 is equal te

S=∑_n=1^10(sin2nπ11 cos2nπ11) nbsp; =(sin2π11 cos2π11)+(sin4π11 cos4π11)+(sin6π11 cos6π11)+⋯+(sin20π11 cos20π11) =(sin2π11+sin4π11+⋯+sin20π11) (cos2π11+cos4π1 ...

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Standard XII

Mathematics

Sum of Cosines of Angles in Arithmetic Progression

The value of ...

Question

The value of $10 \sum n = 1 (sin \frac{2 n π}{11} - cos \frac{2 n π}{11})$ is equal to

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Solution

$S = 10 \sum n = 1 (sin \frac{2 n π}{11} - cos \frac{2 n π}{11})$
$= (sin \frac{2 π}{11} - cos \frac{2 π}{11}) + (sin \frac{4 π}{11} - cos \frac{4 π}{11}) + (sin \frac{6 π}{11} - cos \frac{6 π}{11}) + \dots + (sin \frac{20 π}{11} - cos \frac{20 π}{11})$
$= (sin \frac{2 π}{11} + sin \frac{4 π}{11} + \dots + sin \frac{20 π}{11}) - (cos \frac{2 π}{11} + cos \frac{4 π}{11} + \dots + cos \frac{20 π}{11})$
$= \frac{sin π \cdot sin \frac{10 π}{11}}{sin \frac{π}{11}} - \frac{cos π \cdot sin \frac{10 π}{11}}{sin \frac{π}{11}}$
$= 0 + \frac{sin (π - \frac{π}{11})}{sin \frac{π}{11}}$
$= 1$

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The value of 10∑n=1(sin2nπ11−cos2nπ11) is equal to

S=10∑n=1(sin2nπ11−cos2nπ11) =(sin2π11−cos2π11)+(sin4π11−cos4π11)+(sin6π11−cos6π11)+⋯+(sin20π11−cos20π11) =(sin2π11+sin4π11+⋯+sin20π11)−(cos2π11+cos4π11+⋯+cos20π11) =sinπ⋅sin10π11sinπ11−cosπ⋅sin10π11sinπ11 =0+sin(π−π11)sinπ11 =1

The value of $10 \sum n = 1 (sin \frac{2 n π}{11} - cos \frac{2 n π}{11})$ is equal to