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

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

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