k = 1 ∏ 1 0 0 ( 1 + 2 cos ( 3 1 0 0 + 1 2 π ⋅ 3 k ) ) = ?
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Can you derive the fact?
using componendo dividendo
At some point the cos function is gonna go to zero, so we end up with the product of ones which gives an answer of 1 .
The above product can be split into the sum of two products. Now, for the product of cosines, we know for some k , we are going to have the value of c o s = 0 , then the whole product is 0 too because there is one term which is 0 . The remaining is product of 1 s which is 1
Can you please justify your statement?
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For a hint use the fact that (2cos2A +1) = sin(3A)/sinA.