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

1. 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???

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

Finding the exact value of $\displaystyle \cos 36$ is the same as finding the same value of $\displaystyle \cos \frac{\pi}{5}=\cos t$, because $\displaystyle \frac{180}{5}=36$.

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

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

But we can also do this:

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

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

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

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

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

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

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

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

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

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

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

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

PS: I think this thread should be here.