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Question

3k=1cos2((2k1)π12)=

A
0
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B
12
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C
12
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D
32
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Solution

The correct option is C 32
cos2π12+cos23π12+cos25π12=cos2π12+cos2π4+cos25π12=cos2π12+cos2(π2π12)+(12)2=cos2π12+sin2π12+12=1+12=32

Therefore, Answer is 32

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