Hello I have been tasked with proving the following: [tex]cos(\frac{2 \pi}{5}) = \frac{\sqrt{5} + 1}{4}[/tex] Any hints/idears on how I go about doing that? Sincerely Yours Fred
Start with the unit circle. Then construct a right triangle inside it with one of its angles equal to 2pi/5. From there, use trigonometry and geometry to figure out how to arrive to the answer.
I used Eulers formula [tex]cos(v) = \frac{e^{iv} + e^{-iv}}{2}[/tex] if v = [tex]\frac{2 \pi}{5}[/tex] cos(v) = [tex]\frac{\sqrt{5}-1}{4}[/tex] Is that a wrong approach? /Fred
?? How dou you arrive to that result from euler's formula? From it we just have cos(2pi/5)= (e^(i2pi/5) + e^(-i2pi/5))/2 = cos(2pi/5), because e^(iv)=cos(v)+i sen(v). ??
Using this formula proofs the result doesn't? Cause if I insert the numbers into the formula I get the cos(2pi/5). /Fred
If you plug v=2pi/5 in the following cos(v)=(e^(iv)+e^(-iv))/2 and use then the euler formula you get cos(2pi/5)=cos(2pi/5), not a specific numerical value... How do you arrived to it then? post the steps you have followed.
Sure, I'm sure show the following cos(2 pi / 5) = (sqrt(5) -1)/4 Then I take Eulers formula [tex]cos(v) = \frac{e^{iv} + e^{-iv}}{2}[/tex] and I insert v = (2 pi)/5 into euler which gives (sqrt(5) -1)/4. Thereby cos(2 pi / 5) = (sqrt(5) -1)/4 Is that approach correct? /Fred
If you start with what you are trying to prove as known, then you're not proving nothing. substituting 2pi/5 in the identity your using for cos(v) gives cos(2pi/5)= (e^(i2pi/5)+ e^(-i2pi/5))/2=cos(2pi/5). Which does not give the numerical value you are requested to obtain. You need to evaluate cos(2pi/5) and prove that the result is (sqrt(5)-1)/4. Now, the most elementary way(in which i can think right now) that do this is using the unitary circle and geometry/trigonometry as i mention in my first post.Of course it depends, From what course this problem came? It is likely that you have to use the methods you have learnt from this class. What do yo mean with Gauss proof?
You have to draw the corresponding right triangle inside the circle and then use the pythagoras theorem and other trigonometry concepts to solve the problem. If you want a 'faster', 'algebraic' solution you may prefer to use the results/methods presented here http://mathworld.wolfram.com/TrigonometryAnglesPi5.html and here http://mathworld.wolfram.com/Multiple-AngleFormulas.html You just need to figure out which of them are the ones you need to construct your proof. They do it for sin(pi/5), you need to do it for cos(2pi/5).
Sounds suspiciously like homework, but try using the identity: [tex](\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)[/tex] (which you can easily prove by induction). You can then equate the real parts and substitute [tex]1-\cos^2x[/tex] for [tex]\sin^2x[/tex] to get a quintic polinoleum in [tex]\cos(\frac{2\pi}{5})[/tex] which should be easier for you to factorize than it was for Gauss because you already know the answer. You can then throw away all but one solution because they obviously don't fit. By the way if this is illegible it's because it's the first time I've used tex and I can't get it to display in the preview.