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Math Help - cos5a Expansion

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    cos5a Expansion

    Show that \cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a).

    Thanks in advance.
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    Hi =)

    Quote Originally Posted by Air View Post
    Show that \cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a).

    Thanks in advance.
    I'd suggest you do this way :

    \cos 5a=\cos (4a+a)=\cos 4a \sin a-\sin 4a \cos a=\cos (2*2a) \sin a-\sin(2*2a) \cos a

    Then, use the appropriate formulae :

    \cos 2x=2 \cos^2x

    \sin 2x=2 \cos x \sin x

    \sin^2x=1-\cos^2x



    -------------------

    Or you can start from 16\cos^5a - 20\cos^3a + 5\cos(a)

    And expand while substituting : \cos a=\frac{e^{ia}+e^{-ia}}{2}
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  3. #3
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    Quote Originally Posted by Air View Post
    Show that \cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a).

    Thanks in advance.
    Some smooth approaches have already been given. Here's yet another approach (but only if you're familiar with complex numbers):

    \cos (5a) = Re \left[ e^{i(5a)}\right] = Re \left[ \left( e^{ia} \right)^5 \right] = Re \left[ (\cos (a) + i \sin (a) )^5 \right] .....
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  4. #4
    Junior Member rednest's Avatar
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    Cool

    Quote Originally Posted by Air View Post
    Show that \cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a).

    Thanks in advance.
    For FP3, use De Moivre's Theorem. z^n = r^n[cos\theta+isin\theta]^n = r^n[cos(n\theta)+isin(n\theta)]

    Now binomial theorem gives [cos\theta+isin\theta]^5 =

    [cos^5\theta+5cos^4\theta(i)sin\theta+10cos^3\theta  (i^2)sin^2\theta+
    10cos^2\theta(i^3)sin^3\theta+5cos\theta(i^4)sin^4  \theta+(i^5)sin^5\theta]

    But you know that i^2=-1, i^3=-i, i^4=1, i^5=i and so on

    Considering only the real part of the expansion

    Re[cos\theta+isin\theta]^5 = [cos^5\theta - 10 cos^3\theta sin^2\theta + 5 cos\theta sin^4\theta]

    Consider
    sin^2\theta = 1 - cos^2\theta
    sin^4\theta = (1 - cos^2\theta)^2

    [cos^5\theta - 10 cos^3\theta (1 - cos^2\theta) + 5 cos\theta (1 - cos^2\theta)^2] = [16cos^5\theta - 20 cos^3\theta + 5cos\theta]
    (sorry I ran out of time there, working should be easy)
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