Evaluate cos⁡θcos⁡2θcos⁡3θ+cos⁡2θcos⁡3θcos⁡4θ+… upto n terms

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Evaluate  cos⁡θcos⁡2θcos⁡3θ+cos⁡2θcos⁡3θcos⁡4θ+…  upto n terms

Steve Hirano · Accepted Answer

cos⁡(n−1)t⋅cos⁡nt⋅cos⁡(n+1)t =cos⁡nt(cos⁡2t+cos⁡2nt)2 =cos⁡2tcos⁡nt2+cos⁡nt+cos⁡3nt4 =2cos⁡2t+14⋅cos⁡nt+cos⁡3nt4 Use ∑cos when angles are in arithmetic progression

Elois Puryear · Accepted Answer

Let&#039;s write cn=cos⁡(nθ),sn=sin⁡(nθ) to simplify notation. We observe:cncn+1cn+2=cn+1(cn+1c1+sn+1s1)(cn+1c1−sn+1s1)=cn+1(c{n+1}2c12−s{n+1}2s12)=cn+1(c{n+1}2(c12+s12)−s12)=c{n+1}3−cn+1s1Hence, we see that∑k=1nckck+1ck+2=∑k=2nck3−s1∑k=2nckYou can find a reference for the latter sum here. For the first one, we may use the complex exponential representationcos⁡(θ)=12eiθ+e−iθin order to deduce thatcos⁡(kθ)3=18e3ikθ+3eikθ+3e−ikθ+e−3ikθSumming this is now just a sum of four different geometric series, which should be easy to do (of course, you want to convert it back into sines and cosines at the end). This method can also be used for the second sum, actually.

alenahelenash · Accepted Answer

hint cos⁡((n−1)t)cos⁡(nt)cos⁡((n+1)t)= 18(ei(n−1)t+e−i(n−1)t)(eint+e−int)(ei(n+1)t+e−i(n+1)t) you get then geometric series.

Evaluate \cos \theta \cos 2 \theta \cos 3\theta+\cos 2\theta \cos3\theta \cos

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