Question Number 96907 by bemath last updated on 05/Jun/20
$$\underset{\mathrm{z}\:=\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{cos}\:^{\mathrm{3}} \left(\frac{\pi\mathrm{z}}{\mathrm{3}}\right)\:=\:? \\ $$
Commented by john santu last updated on 05/Jun/20
![cos 3x = 4cos^3 x−3cos x cos^3 x = (1/4) [ cos 3x+3cos x ] Σ_(z = 0) ^(10) (1/4)[ cos 3(((πz)/3)) + 3cos (((πz)/3))] =(1/4) Σ_(z = 0) ^(10) [cos πz + 3cos (((πz)/3))] = (1/4)[ Σ_(z = 0) ^(10) cos πz + 3Σ_(z = 0) ^(10) cos (((πz)/3)) ] =(1/4)[ 1+3 ×((cos (((5π)/3)).sin (((11π)/6)))/(sin ((π/6)))) ] = (1/4)[ 1+(−(3/2))] = −(1/8)](https://www.tinkutara.com/question/Q96908.png)
$$\mathrm{cos}\:\mathrm{3x}\:=\:\mathrm{4cos}\:^{\mathrm{3}} \mathrm{x}−\mathrm{3cos}\:\mathrm{x} \\ $$$$\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\:\mathrm{cos}\:\mathrm{3x}+\mathrm{3cos}\:\mathrm{x}\:\right] \\ $$$$\underset{\mathrm{z}\:=\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{cos}\:\mathrm{3}\left(\frac{\pi\mathrm{z}}{\mathrm{3}}\right)\:+\:\mathrm{3cos}\:\left(\frac{\pi\mathrm{z}}{\mathrm{3}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\underset{\mathrm{z}\:=\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\left[\mathrm{cos}\:\pi\mathrm{z}\:+\:\mathrm{3cos}\:\left(\frac{\pi\mathrm{z}}{\mathrm{3}}\right)\right]\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\:\underset{\mathrm{z}\:=\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{cos}\:\pi\mathrm{z}\:+\:\mathrm{3}\underset{\mathrm{z}\:=\:\mathrm{0}} {\overset{\mathrm{10}} {\sum}}\:\mathrm{cos}\:\left(\frac{\pi\mathrm{z}}{\mathrm{3}}\right)\:\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{1}+\mathrm{3}\:×\frac{\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{3}}\right).\mathrm{sin}\:\left(\frac{\mathrm{11}\pi}{\mathrm{6}}\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}\right)}\:\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}\left[\:\mathrm{1}+\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)\right]\:=\:−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$ \\ $$