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

Question

15929 views around the world

You can reuse this answer
Creative Commons License

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

How do you find the exact value of cos54cos54?