Help with Proof: Sum of Cosines Up to n Terms

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SUMMARY

The discussion centers on the mathematical proof for the sum of cosines up to n terms, specifically represented by the equation $$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$. Participants utilized the substitution $$z=exp(ix)$$ to derive the result, leading to the expression $$(1+z)^n = 2^n cos^n (x/2) cos (nx/2)$$. The final answer is confirmed as the real part of the derived expression, showcasing a clear method for solving the problem.

PREREQUISITES
  • Understanding of complex numbers and Euler's formula
  • Familiarity with binomial coefficients, denoted as $$nCk$$
  • Knowledge of trigonometric identities, particularly cosine
  • Experience with mathematical proofs and manipulation of series
NEXT STEPS
  • Study the derivation of Euler's formula and its applications in trigonometry
  • Learn about binomial expansions and their relevance in combinatorial mathematics
  • Explore the properties of cosine functions and their sums
  • Investigate the use of complex numbers in solving real-valued problems
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Mathematicians, physics students, and anyone interested in advanced trigonometric identities and their proofs will benefit from this discussion.

shen07
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$$\sum_{k=0}^{k=n}(nCk * cos(kx)) = cos(nx/2)*(2cos(x/2))^n$$
 
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I also know that i have to use

$$
z=exp(ix)
$$
$$
(1+z)^n = 2^n cos^n (x/2) cos (nx/2) $$
 
Got the answer..
 
shen07 said:
Got the answer..

Hi shen07 (Wave),

Welcome to MHB! Sorry we couldn't help you quickly enough this time. I'm sure that in the future you'll find guidance with something you are stuck on. Care to share your answer so others may see it?

Jameson
 
The answer could be obtained by choosing the real part after substituting $z=e^{ix}$

Hence we have

$$\Large (1+e^{ix})^n= e^{\frac{ixn}{2}}\left(e^{\frac{-ix}{2}}+e^{\frac{ix}{2}}\right)^n= 2^n e^{\frac{ixn}{2}}\cos^n \left(\frac{x}{2}\right)$$

Clearly the answer is the real part of the above expression .
 

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