# I dont even know how to describe that... :S finding an exact value of cos36??

• Feb 10th 2013, 10:51 PM
Jordi1986
I dont even know how to describe that... :S finding an exact value of cos36??
This one says: Noting that t=pi/5 satisfies 3t=pi-2t , fins the exact value of cos36.....This identities may be uself: cos^2t+sin^2t=1 , sin2t=2sintcost sin3t=3sint-4sin^3t Any clues???
• Feb 11th 2013, 07:00 AM
veileen
Re: I dont even know how to describe that... :S finding an exact value of cos36??
Finding the exact value of $\cos 36$ is the same as finding the same value of $\cos \frac{\pi}{5}=\cos t$, because $\frac{180}{5}=36$.

You are given formulae for sin, so we're going to use them:

$3t=\pi-2t\Rightarrow \sin 3t=\sin (\pi-2t)=\sin 2t=2\sin t \cos t$

But we can also do this:

$\sin 3t=3 \sin t - 4\sin^3 t$

Since $\sin 3t$ is the same no matter how we write it:

$2\sin t \cos t=3 \sin t - 4\sin^3 t$

$t \in \left ( 0, \frac{\pi}{2} \right )\Rightarrow \sin t\neq 0$, that means we can simplify the relation by $\sin t$.

$2 \cos t=3 - 4\sin^2 t \Rightarrow 2\cos t=3-4(1-cos^2 t)$

$\Rightarrow 2\cos t=3-4+4\cos^2 t \Rightarrow 4 \cos t^2-2\cos t-1=0$

Let $x$ be $\cos t$, so $x>0$.

$4 \cos t^2-2\cos t-1=0 \Leftrightarrow 4x^2-2x-1=0$

$\Delta =(-2)^2-4\cdot 4\cdot (-1)=4+16=20$

$x_1=\frac{2+\sqrt{20}}{8}=\frac{2+2\sqrt{5}}{8}= \frac{1+\sqrt{5}}{4}>0$

$x_2=\frac{2-\sqrt{20}}{8}=\frac{2-2\sqrt{5}}{8}=\frac{1-\sqrt{5}}{4} < 0$

So $x=\cos t =\frac{1+\sqrt{5}}{4}$.

PS: I think this thread should be here.