How do you find the exact value of #cos54#?

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Answer 1 · 2016-06-24T06:17:39

#cos54^o=sqrt(10-2sqrt5)/4#

#cos54^o=cos(90^o-36^o)=sin36^o#

Let #A=36^o# hence #5A=180^o#

or #3A=180^o-2A# and hence

#sin3A=sin(180^o-2A)=sin2A# and expanding it we get

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

as #sinA!=0#, dividing by #sinA#, we get

#3-4sin^2A=2cosA# or

#3-4(1-cos^2A)-2cosA=0# or

#4cos^2A-2cosA-1=0#

and #cosA=(2+-sqrt(2^2-4*4*(-1)))/(2*4)=(2+-sqrt20)/8=(1+-sqrt5)/4#

As #cosA# cannot be negative, #cosA=(1+sqrt5)/4# and

#sinA=sqrt(1-(1+sqrt5)^2/16)=sqrt((16-(6+2sqrt5))/16)#

= #sqrt((10-2sqrt5)/16)=sqrt(10-2sqrt5)/4#

Hence #cos54^o=sqrt(10-2sqrt5)/4#