Geometrical Progression
Level
pending
1 + cos (4x) + cos(8x) + cos(12x)....+ (cos nx)
1+cos2x+cos4x+…+cos2(n−1)x=cos2nx−cos3(n−1)x+cos6x−11(cos4x−1)
None of the above
1+cos5x+cos8x+…+cos5(n−1)x=cos5nx−cos4(n−1)x+cos5x−12(cos4x−1)
1+cos4x+cos8x+…+cos4(n−1)x=cos4nx−cos4(n−1)x+cos4x−12(cos4x−1)
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1 solution
1+cos4x+cos8x+…+cos4(n−1)x=R(1+(cos4x+isin4x)+…+ (cos4(n−1)+isin4(n−1)x)=R(1+e 4ix +e 8ix +…+e 4(n−1)ix ) This is a sum of a geometrical progression with a ratio e 4ix . Now sum it. A formula will give an answer: R(1+e 4ix +e 8ix +…+e 4(n−1)ix )=R(e 4nix −1e 4ix −1 ) Now try to give a fraction in another form: e 4nix −1e 4ix −1 =e 4(n−1)ix −e 4nix −e −4ix +12(1−cos4x)
An now it is very easy to take a real part of the new fraction. So, the answer will be: 1+cos4x+cos8x+…+cos4(n−1)x=cos4nx−cos4(n−1)x+cos4x−12(cos4x−1)