How do you find the exact values of cos 4pi/5?

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Answer 1 · 2016-06-09T16:55:21

#=-(sqrt5+1)/4#

We are to evalute
#cos((4pi)/5)=cos(pi-pi/5)=-cos(pi/5)#

This problem of evaluating #cos(pi/5)#can be solved by evaluating #sin(pi/10)# in the following way.

Let #A=pi/10#

#=>5A=pi/2#

#=>3A=pi/2-2A#

#:.cos(3A)=cos(pi/2-2A)#

#=>4cos^3A-3cosA=sin2A=2sinAcosA#

#=>4cos^2A-3=2sinA#

#=>4-4sin^2A-3=2sinA#

#=>4sin^2+2sinA-1=0#

#=>sinA=(-2+sqrt(2^2-4*4* (-1)))/(2*4)#

#=(-2+sqrt20)/8=(sqrt5-1)/4#

#:.sin(pi/10)=(sqrt5-1)/4#,

Now

#cos((4pi)/5)=cos(pi-pi/5)=-cos(pi/5)#

#=-(1-2sin^2(pi/10))#

#=2sin^2(pi/10)-1#

#=2*((sqrt5-1)/4)^2-1#

#=2/16(5+1-2sqrt5)-1#

#=1/8*2*(3-sqrt5)-1#

#=(3-sqrt5)/4-1#

#=(3-sqrt5-4)/4#

#=-(sqrt5+1)/4#