If cos2-3-cos-cos4-3-cos5-3-cos 2-cos7-3-cos40-3-cos41-3-k- then the value of -6k- is

K = cos2π3 cos(π)+cos4π3 cos5π3+cos (2π) cos7π3+⋯+cos40π3 cos41π3 nbsp; = cos2π3+cos4π3+cos (2π)+⋯+cos40π3 (cos(π)+cos5π3+cos7π3+⋯+cos41π3) nbsp; =∑_n= ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If cos2π3-cos...

Question

If $cos \frac{2 π}{3} - cos (π) + cos \frac{4 π}{3} - cos \frac{5 π}{3} + cos (2 π) - cos \frac{7 π}{3} + \dots + cos \frac{40 π}{3} - cos \frac{41 π}{3} = k$ , then the value of $| 6 k |$ is

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Solution

$k = cos \frac{2 π}{3} - cos (π) + cos \frac{4 π}{3} - cos \frac{5 π}{3} + cos (2 π) - cos \frac{7 π}{3} + \dots + cos \frac{40 π}{3} - cos \frac{41 π}{3}$
$= cos \frac{2 π}{3} + cos \frac{4 π}{3} + cos (2 π) + \dots + cos \frac{40 π}{3} - (cos (π) + cos \frac{5 π}{3} + cos \frac{7 π}{3} + \dots + cos \frac{41 π}{3})$
$= 19 \sum n = 0 cos (\frac{2 π}{3} + cos \frac{2 n π}{3}) - 19 \sum n = 0 cos (π + cos \frac{2 n π}{3})$
$= \frac{sin (\frac{80 π}{3}) cos (\frac{2 π}{3} + \frac{19 π}{3})}{sin (\frac{π}{3})} - \frac{sin (\frac{80 π}{3}) cos (π + \frac{19 π}{3})}{sin (\frac{π}{3})}$
$= \frac{sin (\frac{π}{3}) cos (7 π)}{sin (\frac{π}{3})} - \frac{sin (\frac{π}{3}) cos (\frac{2 π}{3})}{sin (\frac{π}{3})}$
$= - 1 - (- \frac{1}{2})$
$= - \frac{1}{2}$
$∴ | 6 k | = 3$

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If cos2π3−cos(π)+cos4π3−cos5π3+cos(2π)−cos7π3+⋯+cos40π3−cos41π3=k, then the value of |6k| is

If $cos \frac{2 π}{3} - cos (π) + cos \frac{4 π}{3} - cos \frac{5 π}{3} + cos (2 π) - cos \frac{7 π}{3} + \dots + cos \frac{40 π}{3} - cos \frac{41 π}{3} = k$ , then the value of $| 6 k |$ is