Question

prove the following...​

Answer 1

Answer:

1+cos⁡θ+cos⁡2θ+...+cos⁡nθ=ℜ(1+cos⁡θ+isin⁡θ+cos⁡2θ+isin⁡2θ+...+cos⁡nθ++isin⁡nθ)=ℜ(1+eiθ+e2iθ+...+eniθ)=ℜ(e(n+1)iθ−1eiθ−1)=ℜ(e(n+1)2iθeiθ2⋅e(n+1)2iθ−e−(n+1)2iθeiθ2−e−iθ2)=ℜ(e(n+1)2iθeiθ2⋅sin⁡(n+1)θ2sin⁡θ2)=ℜ(eiθn2⋅sin⁡(n+1)θ2sin⁡θ2)=cos⁡nθ2⋅sin⁡(n+1)θ2sin⁡θ2=12sin⁡(nθ2+(n+1)θ2)−sin⁡(nθ2−(n+1)θ2)sin⁡θ2=12sin⁡((n+12)θ)+sin⁡θ2sin⁡θ2

Answer 2

see solution in image ......

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