cos theta + cos 3 theta +cos 5 theta+--------cos(2n-1)theta. solve Share with your friends Share 0 Ajanta Trivedi answered this the given expression is: cosθ+cos3θ+cos5θ+.........+cos(2n-1)θ= real part of (cosθ+isinθ)+(cos3θ+isin3θ)+(cos5θ+isin5θ)+...........+cos(2n-1)θ+isin(2n-1)θ=real part of eiθ+ei3θ+ei5θ+...........+ei(2n-1)θ=real part of eiθ.1-ei2nθ1-ei2θ [sum of GP with ratio ei2θ]=real part of 1-ei2nθe-iθ-eiθ=real part of 1-(cos2nθ+isin2nθ)(cosθ-isinθ)-(cosθ+isinθ)=real part of 1-cos2nθ-isin2nθ-2isinθ=real part of i(1-cos2nθ)+sin2nθ2sinθ [multiplying by i to the Nr and Dr=sin2nθ2sinθ hope this helps you 0 View Full Answer
