Question #acd9c

1 Answer
Aug 16, 2017

LHS of given identity

#=sectheta+sec2theta+sec4theta#

#=sec((2pi)/7)+sec((4pi)/7)+sec((8pi)/7)#

#=1/cos((2pi)/7)+1/cos((4pi)/7)+1/cos(pi+pi/7)#

#=1/cos((2pi)/7)+1/cos((4pi)/7)-1/cos(pi/7)#

#=(cos((4pi)/7)cos(pi/7)+cos((2pi)/7)cos(pi/7)-cos((4pi)/7)cos((2pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(2cos((4pi)/7)cos(pi/7)+2cos((2pi)/7)cos(pi/7)-2cos((4pi)/7)cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(cos((5pi)/7)+cos((3pi)/7)+cos((3pi)/7)+cos(pi/7)-cos((6pi)/7)-cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(cos(pi-(2pi)/7)+cos((3pi)/7)+cos((3pi)/7)+cos(pi/7)-cos(pi-pi/7)-cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(2cos((3pi)/7)+2cos(pi/7)-2cos((2pi)/7))/(2cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(cos((3pi)/7)+cos(pi/7)-cos((2pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=-(cos((2pi)/7)-cos(pi/7)-cos((3pi)/7))/(cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

Now the numerator

#cos((2pi)/7)-cos((3pi)/7)-cos(pi/7)#

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

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((4pi)/7)+sin((2pi)/7)-sin((2pi)/7))#

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin(pi-(3pi)/7))#

#=1/(2sin(pi/7))(sin((3pi)/7)-sin(pi/7)-sin((3pi)/7))#

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

#=-1/2#

And the denominator

#=(cos((2pi)/7)cos((4pi)/7)cos(pi/7))#

#=(8sin(pi/7)cos((2pi)/7)cos((4pi)/7)cos(pi/7))/(8sin(pi/7))#

#=(4sin((2pi)/7)cos((2pi)/7)cos((4pi)/7))/(8sin(pi/7))#

#=(2sin((4pi)/7)cos((4pi)/7))/(8sin(pi/7))#

#=sin((8pi)/7)/(8sin(pi/7))#

#=sin(pi+pi/7)/(8sin(pi/7))#

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

So the whole LHS

#=-(-1/2)/(-1/8)=-4=RHS#