The value of ∑13k=11sin(π4+(k−1)π)6)sin(π4+kπ6) is equal to
A
3−√3
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B
2(3−√3)
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C
2(√3−1)
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D
2(2−√3)
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Solution
The correct option is C2(√3−1) The value of ∑13k=11sin(π4+(k−1)π6)sin(π4+kπ6)∑13k=12sinπ6[sin{π4+kπ6−(π4+(k−1)π6)}sin(π4+(k−1)π6))sin(π4+kπ6)]∑13k=12[cot(π4+(k−1)π6)−cot(π4+kπ6)]=2[{cotπ4−cot(π4+π6)}+{cot(π4+π6)−cot(π4+2π6)}+…+{cot(π4+12π6)−cot(π4+13π6)}]=2[cotπ4−cot(π4+13π6)]=2[1−cot5π12]=2[1−√3−1√3+1]=2[1−(2−√3)]=2(√3−1)