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Question Number 140750 by SOMEDAVONG last updated on 12/May/21

Σ_(k=0) ^n cos^3 k=?

$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{cos}^{\mathrm{3}} \mathrm{k}=? \\ $$

Answered by Dwaipayan Shikari last updated on 12/May/21

4cos^3 k−3cosk=cos3k⇒cos^3 k=(1/4)(3cosk+cos3k) Σ_(k=0) ^n cos^3 k=1+(3/4)Σ_(k=1) ^n cos(k)+(1/4)Σcos(3k) Now Σ_(n=1) ^n cos(nx)=((cos(((n+1)/2)x)sin(((nx)/2)))/(sin((x/2)))) So the sum becomes 1+(3/4)(((cos(((n+1)/2))sin((n/2)))/(sin((1/2))))+((cos(3((n+1)/2))sin(((3n)/2)))/(3sin((1/2))))) =1+(1/4)(((3cos(((n+1)/2))sin((n/2))+cos(((3n+3)/2))sin(((3n)/2)))/(sin((1/2)))))

$$\mathrm{4}{cos}^{\mathrm{3}} {k}−\mathrm{3}{cosk}={cos}\mathrm{3}{k}\Rightarrow{cos}^{\mathrm{3}} {k}=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{cosk}+{cos}\mathrm{3}{k}\right) \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{cos}^{\mathrm{3}} {k}=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({k}\right)+\frac{\mathrm{1}}{\mathrm{4}}\Sigma{cos}\left(\mathrm{3}{k}\right) \\ $$$${Now}\:\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}\left({nx}\right)=\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}\right){sin}\left(\frac{{nx}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}\: \\ $$$${So}\:{the}\:{sum}\:{becomes}\:\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{{n}}{\mathrm{2}}\right)}{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}+\frac{{cos}\left(\mathrm{3}\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{3}{n}}{\mathrm{2}}\right)}{\mathrm{3}{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}{cos}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right){sin}\left(\frac{{n}}{\mathrm{2}}\right)+{cos}\left(\frac{\mathrm{3}{n}+\mathrm{3}}{\mathrm{2}}\right){sin}\left(\frac{\mathrm{3}{n}}{\mathrm{2}}\right)}{{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\right) \\ $$

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