Given #cos(2pi/5) = (sqrt(5)-1)/4#, what is cos(3pi/5)?

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Answer 1 · 2016-03-10T17:09:13

#(1-sqrt(5))/4#

#cos (theta)= -cos(pi-theta)#

therefore

#cos(3pi/5)=cos (pi-2pi/5)=-cos(2pi/5)#

#=(1-sqrt(5))/4#

Answer 2 · 2016-03-10T17:20:24

#=-(sqrt5-1)/4#

#cos( (3pi)/5)=cos (pi-(2pi)/5)=-cos( (2pi)/5)=-(sqrt5-1)/4#