How do you evaluate $cos (\frac{6 π}{5})$ $cos((6pi)/5)$ ?

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Answer 1 · 2017-03-25T15:28:06

$cos (\frac{6 π}{5})$ $cos((6pi)/5)$

$= cos (π + \frac{π}{5})$ $=cos(pi+pi/5)$

$= - cos (\frac{π}{5})$ $=-cos( pi/5)$

Now let

$θ = \frac{π}{5}$ $theta=pi/5$

$\Rightarrow 5 θ = π$ $=>5theta=pi$

$\Rightarrow 3 θ = π - 2 θ$ $=>3theta=pi-2theta$

$\Rightarrow sin 3 θ = sin (π - 2 θ)$ $=>sin3theta=sin(pi-2theta)$

$\Rightarrow sin 3 θ = sin 2 θ$ $=>sin3theta=sin2theta$

$\Rightarrow 3 sin θ - 4 {sin}^{3} θ = 2 sin θ cos θ$ $=>3sintheta-4sin^3theta=2sinthetacostheta$

$\Rightarrow sin θ (3 - 4 {sin}^{2} θ) = 2 sin θ cos θ$ $=>sintheta(3-4sin^2theta)=2sinthetacostheta$

$\Rightarrow (3 - 4 {sin}^{2} θ) = 2 cos θ$ $=>(3-4sin^2theta)=2costheta$ as $sin θ = sin (\frac{π}{5}) \neq 0$ $sintheta =sin(pi/5)!=0$

$\Rightarrow (3 - 4 + 4 {cos}^{2} θ) = 2 cos θ$ $=>(3-4+4cos^2theta)=2costheta$

$\Rightarrow 4 {cos}^{2} θ - 2 cos θ - 1 = 0$ $=>4cos^2theta-2costheta-1=0$

So

$cos θ = \frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \times 4 (- 1)}}{2 \times 4}$ $costheta=(2pmsqrt((-2)^2-4xx4(-1)))/(2xx4)$

$cos θ = \frac{2 \pm 2 \sqrt{5}}{2 \times 4}$ $costheta=(2pm2sqrt5)/(2xx4)$

$cos θ = \frac{1 \pm \sqrt{5}}{4}$ $costheta=(1pmsqrt5)/4$

as $cos θ = \frac{1 - \sqrt{5}}{4} < 0 not possible$ $costheta=(1-sqrt5)/4<0" not possible"$

$cos θ = \frac{1 + \sqrt{5}}{4}$ $costheta=(1+sqrt5)/4$

$\Rightarrow cos (\frac{π}{5}) =$ $=>cos(pi/5)=$

So

$cos (\frac{6 π}{5}) = - cos (\frac{π}{5}) = - \frac{1 + \sqrt{5}}{4}$ $cos((6pi)/5)=-cos(pi/5)=-(1+sqrt5)/4$

How do you evaluate cos((6pi)/5)cos(6π5)cos((6pi)/5)?