Question

prove that costheta×cos2theta×cos3theta=1/4

Answer 1

R(cosθ+isinθ)+(cosθ+isinθ)2+…+(cosθ+isinθ)n

I realized that this is a Geometric Progression, so its in the form:

a+ar+ar2+....+arn , where a=(cosθ+isinθ) and r=(cosθ+isinθ)

So I will apply the formula for the Sum of a G.P to my problem.

R(cosθ+isinθ)(1−(cosθ+isinθ)n)1−(cosθ+isinθ)

I applied the De Movire Theorem and simplified as follows:

Rcosθ+isinθ−cos(n+1)θ+isin(n+1)θ1−(cosθ+isinθ)

Rcosθ+isinθ−cos(n+1)θ+isin(n+1)θ1−(cosθ+isinθ)(1+(cosθ+isinθ))(1+(cosθ+isinθ))

Rcosθ+isinθ−cos(n+1)θ+isin(n+1)θ+cos2θ+2icosθsinθ−sin2θ−cosθcos(n+1)θ−isinθcos(n+1)θ+icosθsin(n+1)θ−sinθsin(n+1)θ1−cos2θ−2icosθsinθ+sin2θ

Rcosθ+isinθ−cos(n+1)θ+isin(n+1)θ+cos(2θ)+isin(2θ)−cos(n+1)θ(cosθ+isinθ)+(icosθ−sinθ)sin(n+1)θ1−cos2θ−2icosθsinθ+sin2θ

Rcosθ+isinθ+cos(2θ)+isin(2θ)−cos(n+1)θ(1+cosθ+isinθ)+(1+icosθ−sinθ)sin(n+1)θ2sin2θ−2icosθsinθ

R[cosθ+isinθ+cos(2θ)+isin(2θ)−cos(n+1)θ(1+cosθ+isinθ)+(1+icosθ−sinθ)sin(n+1)θ]2sinθ(sinθ−icosθ)(sinθ+icosθ)(sinθ+icosθ)

[−cos(n+1)θ(1+cosθ+isinθ)+(1+icosθ−sinθ)sin(n+1)θ]2sinθ(sinθ+icosθ)1

[−cos(n+1)θ(sinθ+cosθsinθ−cosθsinθ)+(sinθ−cos2θ−sin2θ)sin(n+1)θ]2sinθ

[−cos(n+1)sinθ+(sinθ−1)sin(n+1)θ]2sinθ