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\sum _ { k = 1 } ^ { 3 } \cos ^ { 2 } ( ( 2 k - 1 ) \frac { \pi } { 12 } ) |
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Answer» search MATHS avatar Asked on October 15, 2019 by Mantasha Khattri The value of k=1 ∑ 13
sin( 4 π
+ 6 (k−1)π
)sin( 4 π
+ 6 kπ
) 1
is equal to
Share Save VIDEO EXPLANATION ANSWER k=1 ∑ 13
sin 6 π
(sin( 4 π
+ 6 kπ
)sin( 4 π
+(k−1) 6 π
)) sin[( 4 π
+ 6 kπ
)−( 4 π
+(k−1) 6 π
)]
=2 k=1 ∑ 13
(cot( 4 π
+(k−1) 6 π
)−cot( 4 π
+ 6 kπ
)) using compound angle formula. =2(cot 4 π
−cot( 4 π
+ 6 13π
)) =2(1−cot( 12 29π
)) =2(1−cot( 12 5π
)) =2(1−(2− 3
)) =2(−2+ 3
) =2( 3
−1) Step-by-step explanation: follow me OR Mark BRAINLIEST |
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