How do you evaluate this #cos((2pi)/7)cos((3pi)/7)cos((6pi)/7)# ?

Question

Please, explain me EVERYTHING, I am pretty awful at trigonometry

2970 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-22T21:14:06

#cos((2pi)/7)*cos((3pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))*4*2sin((2pi)/7)cos((2pi)/7)*cos((3pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))*2*2sin((4pi)/7)*cos((3pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))*2*2sin(pi-(3pi)/7)*cos((3pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))*2*2sin((3pi)/7)*cos((3pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))*2sin((6pi)/7)*cos((6pi)/7)#

#=1/(8sin((2pi)/7))sin((12pi)/7)#

#=1/(8sin((2pi)/7))sin(2pi-(2pi)/7)#

#=1/(8sin((2pi)/7))(-sin((2pi)/7))#

#=-1/8#