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- #1

- Feb 14, 2012

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I don't see any way to approach this problem but my attempt is to group cos(a).cos(999a), then cos(2a).cos(998a), and so on.

Then I get 500 pairs where the first two pairs are 1/2[cos(100a)+cos(998a)] and 1/2[cos(100a)+cos(996a)], and etc.

Next, I see that cos(a)cos(2a)cos(3a)...cos(999a)= 1/2^500[[cos(100a)+cos(998a)][cos(100a)+cos(996a)][cos(100a)+cos(994a)][...].

Now, I'm stuck. I don't know how to proceed.

Any help, please?