Results 1 to 3 of 3

Thread: Cos 3pi/5

  1. #1
    Member
    Joined
    Oct 2012
    From
    Nowhere
    Posts
    82

    Cos 3pi/5

    I don't understand the step where 2Sin^2(3pi/10) = 2((sqrt5+1)/4) Cos 3pi/5-steps.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    6,602
    Thanks
    1712

    Re: Cos 3pi/5

    Hey Eraser147.

    Try setting up a right angled triangle or using sums or differences formula to get the value.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    Posts
    671
    Thanks
    136

    Re: Cos 3pi/5

    Consider the right-angled triangle with acute angles $\displaystyle \frac{3\pi}{10} $ and $\displaystyle \frac{2\pi}{10} $ and note that $\displaystyle \sin{\left(\frac{3\pi}{10}\right)}=\cos{ \left( \frac{2\pi}{10}\right) }.$

    To ease typing, let $\displaystyle \frac{\pi}{10}=A,$ then $\displaystyle \sin 3A=\cos 2A.$

    Use the standard identities, $\displaystyle \sin 3A =3\sin A-4\sin^{3} A .............(1) \text{ and } \cos 2A= 1-2\sin^{2} A ............(2)$

    and we have the cubic $\displaystyle 4\sin^{3} A -2\sin^{2}A-3\sin A +1 = 0.$

    $\displaystyle (\sin A -1)$ is a factor, remove this and we are left with the quadratic

    $\displaystyle 4\sin^{2} A +2\sin A -1=0, $ and solving this, (for the angle in the $\displaystyle 0,\pi/2$ range),

    $\displaystyle \sin A = \frac{-1+\sqrt{5}}{4}.$

    Substituting this into (1) and simplifying leads to $\displaystyle \sin 3A = \sin{ \ \left( \frac{3\pi}{10} \right)} = \frac{1+\sqrt{5}}{4}.$
    Follow Math Help Forum on Facebook and Google+

Search tags for this page

Search Tags


/mathhelpforum @mathhelpforum