How do you evaluate #cos((6pi)/5)#?

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Answer 1 · 2017-03-25T15:28:06

#cos((6pi)/5)#

#=cos(pi+pi/5)#

#=-cos( pi/5)#

Now let

#theta=pi/5#

#=>5theta=pi#

#=>3theta=pi-2theta#

#=>sin3theta=sin(pi-2theta)#

#=>sin3theta=sin2theta#

#=>3sintheta-4sin^3theta=2sinthetacostheta#

#=>sintheta(3-4sin^2theta)=2sinthetacostheta#

#=>(3-4sin^2theta)=2costheta# as #sintheta =sin(pi/5)!=0#

#=>(3-4+4cos^2theta)=2costheta#

#=>4cos^2theta-2costheta-1=0#

So

#costheta=(2pmsqrt((-2)^2-4xx4(-1)))/(2xx4)#

#costheta=(2pm2sqrt5)/(2xx4)#

#costheta=(1pmsqrt5)/4#

as # costheta=(1-sqrt5)/4<0" not possible"#

#costheta=(1+sqrt5)/4#

#=>cos(pi/5)=#

So

#cos((6pi)/5)=-cos(pi/5)=-(1+sqrt5)/4#