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Math Help - trigonometric expansion

  1. #1
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    Exclamation trigonometric expansion

    Challenge your maths skills...

    Question:
    Express sin13θ in the form
    Asinθ + Bsin^3θ + Csin^5θ +Dsin^7θ + Esin^9θ +Fsin11^θ + Gsin^13θ
    A to G being natural numbers.
    Last edited by mr fantastic; March 17th 2009 at 02:49 AM. Reason: Removed unjustified and potentially insulting comments
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  2. #2
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    Quote Originally Posted by cogito_ergo_sum View Post
    Challenge your maths skills...

    Question:
    Express sin13θ in the form
    Asinθ + Bsin^3θ + Csin^5θ +Dsin^7θ + Esin^9θ +Fsin11^θ + Gsin^13θ
    A to G being natural numbers..

    Its up to you to show us what you got!!
    Help... this is too hard for me...

    i think i need to use complex numbers somewhere... help me, maths buddies!
    You need to use two formulas...

    \sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}

    and

    \sin^2{x} + \cos^2{x} = 1.

    If you do some rearranging of the second, and some expanding of the first, it should fall into place...
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  3. #3
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    Quote Originally Posted by Prove It View Post
    You need to use two formulas...

    \sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}

    and

    \sin^2{x} + \cos^2{x} = 1.

    If you do some rearranging of the second, and some expanding of the first, it should fall into place...
    Can you please explain how to use the first formula... i am a bit confused. tahnks a lot though
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  4. #4
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    Quote Originally Posted by cogito_ergo_sum View Post
    Can you please explain how to use the first formula... i am a bit confused. tahnks a lot though
    Have you ever seen summation notation before?

    Basically it's a sum, and the limits on the sum are your counter.

    So here, n = 13.

    So k will go between 0 and \frac{13 - 1}{2} = 6.

    The \left(_k^{n - k -1}\right) is a term in Pascal's triangle.

    It might help if you did some research on summation (sigma) notation and Pascal's Triangle.

    Summation - Wikipedia, the free encyclopedia

    Pascal's triangle - Wikipedia, the free encyclopedia
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Have you ever seen summation notation before?

    Basically it's a sum, and the limits on the sum are your counter.

    So here, n = 13.

    So k will go between 0 and \frac{13 - 1}{2} = 6.

    The \left(_k^{n - k -1}\right) is a term in Pascal's triangle.

    It might help if you did some research on summation (sigma) notation and Pascal's Triangle.

    Summation - Wikipedia, the free encyclopedia

    Pascal's triangle - Wikipedia, the free encyclopedia

    I did exactly that and it doesn't work out properly. I used the binomial theorem and still stuck.
    Thanks a lot for the help but i still don't know where i went wrong?
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by cogito_ergo_sum View Post
    I did exactly that and it doesn't work out properly. I used the binomial theorem and still stuck.
    Thanks a lot for the help but i still don't know where i went wrong?
    First of all you should thank Prove It that he showed you the way rather than giving you the answer directly and that's the real spirit of MHF

    Now it comes to your question , since you are still unable to catch with it , I will give you an example along with the step numbers**

    example for sin(3\theta)

    First step

    <br />
\sin{(nx)} = \sin{x}\sum_{k = 0}^{\frac{n-1}{2}} (-1)^k \left(_k^{n - k - 1}\right)2^{n - 2k - 1}\cos^{n - 2k - 1}{x}

    Put n = 3 and accordingly k =1 so it becomes



    <br />
\sin{(3x)} = \sin{x}\sum_{k = 0}^{\frac{3-1}{2}} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}

    <br />
  \sin{(3x)} = \sin{x}\sum_{k = 0}^{1} (-1)^k \left(_k^{3 - k - 1}\right)2^{3 - 2k - 1}\cos^{3 - 2k - 1}{x}


    Step 2


    Expand the terms

    <br />
 \sin{(3x)} = \sin{x}[ \{(-1)^0 \left(_0^{3 - 0 - 1}\right)2^{(3 - (2 \times 0) - 1)}\cos^{(3 - (2\times 0) - 1)}{x}\}+ \{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\}]

    ---> \left(_0^{3 - 0 - 1}\right) = \frac{2!}{(2-0)! \times 0!}

    j! ~(spelled ~as~ j~ factorial)~ = j \times (j-1)\times (j-2)\times......1

    Remember that

    1!= 1 = 0!

    So


     \frac{2!}{(2-0)! \times 0!} = 1

    Similarly
    --->
    \left(_1^{3 -1 - 1}\right) = \frac{1!}{0! \times 1!}=1

    Step 3


    Now gather the values of each term

    First term

    <br />
 \{(-1)^0 \left(_0^{3 - 0 - 1}\right)2^{(3 - (2 \times 0) - 1)}\cos^{3 - 2\times 0 - 1}{x}\} = 1 \times 1 \times 4 \times cos^2 (x)  = 4(1-sin^2(x))= 4 - 4sin^2(x)

    Second term
    \{ (-1)^1 \left(_1^{3 -1 - 1}\right)2^{(3 - (2 \times 1) - 1)}\cos^{(3 - (2\times 1) - 1)}{x}\} = -1\times 1 \times 1 \times 1  = -1

    Step 4

    Add First and second term

    3-4sin^2(x)

    Put this in the Step 2 which had those two terms

    <br />
 \sin{(3x)}  = \sin{x}[3-4sin^2(x)] = 3sin(x)-4sin^3(x)

    Thus we have expressed sin(3x) in terms of
    Asin(x) + Bsin^3(x)

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    **Tell the step number you didn't get


    -All you need to do is do the same way for sin(13x) find k, expand,find term values and go ahead
    Last edited by ADARSH; March 16th 2009 at 11:22 PM.
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