13) ans is 1/2
put \theta =\frac{(2n+1)\pi }{11} wer n=0,1,2...
therfore 6\theta =(2n+1)\pi -5\theta
therfore cos6\theta =-cos5\theta
Now
putting cosθ=x
cos6\theta =2cos^2 3\theta -1=2(4 cos^{3}-3cos\theta )^{2}-1=32x^{6}-48x^{4}+18x^{2}+1(a typo-its 4cos3θ)
cos5\theta =16cos^{5}\theta +20cos^3\theta +5cos\theta =16x^5+20x^3+5x
threfore
(x+1)(32x^5-16x^4-42x^3+12x^2+6x-1)=0
now the roots of the given eq r cos(\frac{r\pi }{11}) wer r=1,3,5.....9
so the given cosθ + cos 3θ +cos 5θ + cos 7θ +cos 9θ=sum of the roots of the given eq=-(coefficient of x4)/( coefficient of x5)=1/2
dont confuse this θ with the given θ in the q
1the \ question \ can \ also \ be \ done \ with \ the \ help \ of \ complex \ roots \ of \ unity \\ Consider \\ z^{11} +1 =0 the \ roots \ of \ this \ are \\ -1,z,\bar{z} , z^3 , \bar{z^3}, z^5 , \bar{z^5},z^7 , \bar{z^7}, z^9 , \bar{z^9},\\ \\ dividing \ the \ equation \ by \ z^5 \ we \ get \ \\ (z^5 + \frac{1}{z^5})-(z^4 + \frac{1}{z^4})+(z^3 + \frac{1}{z^3})-(z^2 + \frac{1}{z^2})+(z + \frac{1}{z})-1 \\ now \\ z+\frac{1}{z}= t ,and \ find \\ z^2 + \frac{1}{z^2} ,z^3 + \frac{1}{z^3},z^4 + \frac{1}{z^4},z^5 + \frac{1}{z^5}\ in \ terms \ of \ t \\ the \ polynomial \ on \ t \ turns \ out \ to \ be \\ t^5 - t^4 +6t^3 +3t^2 +6t -1; sum \ of \ roots \ is \ 1 \\ but\ the \ roots \ are \ 2cos\alpha_i \ as, \ z+ \frac{1}{z}=2cos\alpha \\ hence \ answer \ of \ required \ sum \ is \ \\ \frac{1}{2}
1uttara,can u proceed n show for sum 1.........
