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Question Number 11900 by @ANTARES_VY last updated on 04/Apr/17

Calculate.  cos(𝛑/7)Γ—cos((4𝛑)/7)Γ—cos((5𝛑)/7).

$$\boldsymbol{\mathrm{Calculate}}. \\ $$$$\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\pi}}{\mathrm{7}}Γ—\boldsymbol{\mathrm{cos}}\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{7}}Γ—\boldsymbol{\mathrm{cos}}\frac{\mathrm{5}\boldsymbol{\pi}}{\mathrm{7}}. \\ $$

Answered by ajfour last updated on 04/Apr/17

= cos (Ο€/7)cos ((4Ο€)/7)[βˆ’cos (Ο€βˆ’((5Ο€)/7)) ]  = βˆ’cos (Ο€/7)cos ((4Ο€)/7)cos ((2Ο€)/7)  = ((βˆ’1)/(2sin (Ο€/7)))[ 2sin (Ο€/7)cos (Ο€/7)][cos ((2Ο€)/7)cos ((4Ο€)/7) ]  =βˆ’(1/(2sin (Ο€/7)))[sin ((2Ο€)/7)cos ((2Ο€)/7)]cos ((4Ο€)/7)  = βˆ’(1/(4sin (Ο€/7)))(sin ((4Ο€)/7)cos ((4Ο€)/7))  = βˆ’(((sin ((8Ο€)/7))/(8sin (Ο€/7))))= βˆ’((sin (Ο€+(Ο€/7)))/(8sin (Ο€/7)))  = βˆ’((βˆ’sin (Ο€/7))/(8sin (Ο€/7))) = (1/8) .

$$=\:\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\left[βˆ’\mathrm{cos}\:\left(\piβˆ’\frac{\mathrm{5}\pi}{\mathrm{7}}\right)\:\right] \\ $$$$=\:βˆ’\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$$=\:\frac{βˆ’\mathrm{1}}{\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}}\left[\:\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\right]\left[\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\:\right] \\ $$$$=βˆ’\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}}\left[\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{7}}\right]\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}} \\ $$$$=\:βˆ’\frac{\mathrm{1}}{\mathrm{4sin}\:\frac{\pi}{\mathrm{7}}}\left(\mathrm{sin}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{7}}\right) \\ $$$$=\:βˆ’\left(\frac{\mathrm{sin}\:\frac{\mathrm{8}\pi}{\mathrm{7}}}{\mathrm{8sin}\:\frac{\pi}{\mathrm{7}}}\right)=\:βˆ’\frac{\mathrm{sin}\:\left(\pi+\frac{\pi}{\mathrm{7}}\right)}{\mathrm{8sin}\:\frac{\pi}{\mathrm{7}}} \\ $$$$=\:βˆ’\frac{βˆ’\mathrm{sin}\:\left(\pi/\mathrm{7}\right)}{\mathrm{8sin}\:\left(\pi/\mathrm{7}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\:. \\ $$

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