Summation of trigonometric identities

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  1. johnconnor92's Avatar
    1 08-06-2012  03:42
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    Summation of trigonometric identities
    Prove that \displaystyle\sum_{k=0}^n \sin k\theta = \dfrac{\cos (\theta /2) - cos (n + 1/2)\theta}{2 \sin (\theta /2)}

    Attempt:
    \displaystyle\sum_{k=0}^n \sin k\theta = \sin 0 + \sin \theta + \sin 2\theta + ... + \sin (n\theta) \text{by de moivre's theorem,} \displaystyle\sum_{k=0}^n e^{i(k\theta)} = 0 + e^{i(\theta)} + e^{i(2\theta)} +e^{i(3\theta)} + ... + e^{i(n\theta)} \text{by sum of G.P.,} = \dfrac{e^{i(\theta)}(1 - e^{i(n\theta)})}{1- e^{i(\theta)}} = \dfrac{e^{i(\theta)}(1 - e^{i(n\theta)})(1 + e^{i(\theta)}}{(1- e^{i(\theta)})(1+ e^{i(\theta)})}

    But the powers of the exp don't match the target of which I'm supposed to prove. Pointers anyone?
  2. BabyMaths's Avatar
    2 08-06-2012  07:51
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    Re: Summation of trigonometric identities
    (Original post by johnconnor92)
    Prove that \displaystyle\sum_{k=0}^n \sin k\theta = \dfrac{\cos (\theta /2) - cos (n + 1/2)\theta}{2 \sin (\theta /2)}

    Attempt:
    \displaystyle\sum_{k=0}^n \sin k\theta = \sin 0 + \sin \theta + \sin 2\theta + ... + \sin (n\theta) \text{by de moivre's theorem,} \displaystyle\sum_{k=0}^n e^{i(k\theta)} = 0 + e^{i(\theta)} + e^{i(2\theta)} +e^{i(3\theta)} + ... + e^{i(n\theta)} \text{by sum of G.P.,} = \dfrac{e^{i(\theta)}(1 - e^{i(n\theta)})}{1- e^{i(\theta)}} = \dfrac{e^{i(\theta)}(1 - e^{i(n\theta)})(1 + e^{i(\theta)}}{(1- e^{i(\theta)})(1+ e^{i(\theta)})}

    But the powers of the exp don't match the target of which I'm supposed to prove. Pointers anyone?
    e^0=1
  3. johnconnor92's Avatar
    3 08-06-2012  08:28
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    Re: Summation of trigonometric identities
    Gah stupid mistake. Here's a revised one.

    Prove that \displaystyle\sum_{k=0}^n \sin k\theta = \dfrac{\cos (\theta /2) - cos (n + 1/2)\theta}{2 \sin (\theta /2)}

    Attempt:
    \displaystyle\sum_{k=0}^n \sin k\theta = \sin 0 + \sin \theta + \sin 2\theta + ... + \sin (n\theta) \text{by de moivre's theorem,} \displaystyle\sum_{k=0}^n e^{i(k\theta)} = 1 + e^{i(\theta)} + e^{i(2\theta)} +e^{i(3\theta)} + ... + e^{i(n\theta)} \text{by sum of G.P.,} = \dfrac{(1 - e^{i((n+1)\theta)})}{1- e^{i(\theta)}} = \dfrac{(1 - e^{i((n+1)\theta)})(1 + e^{i(\theta)}}{(1- e^{i(\theta)})(1+ e^{i(\theta)})}

    But the powers of the exp don't match the target of which I'm supposed to prove. Pointers anyone?
  4. Allofthem's Avatar
    4 08-06-2012  08:43
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    Re: Summation of trigonometric identities
    Don't they? I've not looked too closely but I think you may have forgotten that e^(ix) = cosx + isinx, so you've summed cos(kx) + i sin(kx), so you ought to be taking just the imaginary part of that sum you've got, that will probably mess about with it suitably well as to make it look similar.
  5. BabyMaths's Avatar
    5 08-06-2012  08:51
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    Re: Summation of trigonometric identities
    You could start like this:

    \sum_{k=0}^n \sin k\theta=\frac{1}{2\sin \frac{1}{2}\theta}\sum_{k=0}^n 2\sin(k\theta)\sin(\frac{1}{2} \theta)
  6. johnconnor92's Avatar
    6 08-06-2012  09:33
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    Re: Summation of trigonometric identities
    (Original post by Allofthem)
    Don't they? I've not looked too closely but I think you may have forgotten that e^(ix) = cosx + isinx, so you've summed cos(kx) + i sin(kx), so you ought to be taking just the imaginary part of that sum you've got, that will probably mess about with it suitably well as to make it look similar.
    I do know about that, but if we look at the powers of the natural constant how are we supposed to get a half out of the terms? I didn't go on because I hesitated on seeing the powers.

    (Original post by BabyMaths)
    You could start like this:

    \sum_{k=0}^n \sin k\theta=\frac{1}{2\sin \frac{1}{2}\theta}\sum_{k=0}^n 2\sin(k\theta)\sin(\frac{1}{2} \theta)
    I know about this method. Tried it and done it, too. But I just want to nail the question with the method here.
  7. DFranklin's Avatar
    7 08-06-2012  10:29
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    Re: Summation of trigonometric identities
    (Original post by johnconnor92)
    = \dfrac{(1 - e^{i((n+1)\theta)})}{1- e^{i(\theta)}}
    As you probably know, You want to take the imaginary part of this, which means you need to make the denominator easier to deal with.

    The normal way of doing this would be to multiply top and bottom by (1-e^{-i\theta}), which makes the denominator real.

    A better method here is to multiply top and bottom by e^{-i\theta/2}, which leaves the denominator as e^{-i\theta/2}-e^{i\theta/2}, which is pure imaginary. It should work out fairly easily from there.
  8. johnconnor92's Avatar
    8 08-06-2012  12:40
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    Re: Summation of trigonometric identities
    (Original post by DFranklin)
    As you probably know, You want to take the imaginary part of this, which means you need to make the denominator easier to deal with.

    The normal way of doing this would be to multiply top and bottom by (1-e^{-i\theta}), which makes the denominator real.

    A better method here is to multiply top and bottom by e^{-i\theta/2}, which leaves the denominator as e^{-i\theta/2}-e^{i\theta/2}, which is pure imaginary. It should work out fairly easily from there.
    OF COURSE! I got the complex conjugate for exponentials wrong! OMGWTFBBQ
  9. DFranklin's Avatar
    9 08-06-2012  13:13
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    Re: Summation of trigonometric identities
    (Original post by johnconnor92)
    OF COURSE! I got the complex conjugate for exponentials wrong! OMGWTFBBQ
    Note what I said about the "better" method here (which doesn't use the complex conjugate). It's a trick worth knowing.
  10. johnconnor92's Avatar
    10 09-06-2012  03:59
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    Re: Summation of trigonometric identities
    (Original post by DFranklin)
    As you probably know, You want to take the imaginary part of this, which means you need to make the denominator easier to deal with. The normal way of doing this would be to multiply top and bottom by (1-e^{-i\theta}), which makes the denominator real.
    \text{multiplying with complex conjugate,} \dfrac{(1 - e^{i((n+1)\theta)})(1 - e^{-i(\theta)}}{(1- e^{i(\theta)})(1- e^{-i(\theta)})} = \dfrac{1-e^{-i\theta}-e^{i(n+1)\theta}+e^{in\theta}}{1-e^{-i\theta} - e^{i\theta} +1 } \text{taking the imaginary parts and solving the denominator,} = \dfrac{1-(\cos\theta +i\sin\theta)+(\cos n\theta + i\sin n\theta) - (\cos (n+1)\theta + i\sin (n+1)\theta)}{2-2\cos \theta} = \dfrac{-\sin\theta+ \sin n\theta -\sin (n+1)\theta)}{4\sin^2(\theta/2)} = \dfrac{-2\cos(\theta/2)\sin(\theta/2)-2\sin(\theta/2) - 2\sin(\theta/2)\cos(n+1/2)\theta}{4\sin^2(\theta/2)} = \dfrac{\cos(\theta/2)+ \cos(n+1/2)\theta}{2\sin(\theta/2)}
    Hmm... Something went wrong.

    A better method here is to multiply top and bottom by e^{-i\theta/2}, which leaves the denominator as e^{-i\theta/2}-e^{i\theta/2}, which is pure imaginary. It should work out fairly easily from there.
    \dfrac{(1 - e^{i((n+1)\theta)})(e^{-i\theta/2})}{(1- e^{i(\theta)})(e^{-i\theta/2})} = \dfrac{e^{-i\theta/2} - e^{i[(n+1)-1/2]\theta}}{e^{-i\theta/2} - e^{i\theta/2}} = \dfrac{\cos(\theta/2) - i\sin(\theta/2) - \cos(n+1/2)\theta +\sin(n+1/2)\theta)}{cos(\theta/2) - i\sin(\theta/2) - i\sin(\theta/2) -\cos(\theta/2)} \text{taking the imaginary parts,} = \dfrac{-\sin(\theta/2)+\sin(n+1/2)\theta}{-2\sin(\theta/2)}

    This is amazing! I always thought denominators with an imaginary numbers cannot be navigated like that of real numbers. But what went wrong in the above calculation? Thank you so much!
    Last edited by johnconnor92; 09-06-2012 at 04:05.
  11. johnconnor92's Avatar
    11 14-06-2012  13:49
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    Re: Summation of trigonometric identities
    (Original post by DFranklin)
    Note what I said about the "better" method here (which doesn't use the complex conjugate). It's a trick worth knowing.
    I proved the summation via the complex conjugate method. Here's another attempt at the pointer you gave me:

    \dfrac{(1 - e^{i((n+1)\theta)})(e^{-i\theta/2})}{(1- e^{i(\theta)})(e^{-i\theta/2})} = \dfrac{e^{-i\theta/2} - e^{i[(n+1)-1/2]\theta}}{e^{-i\theta/2} - e^{i\theta/2}} = \dfrac{\cos(\theta/2) - i\sin(\theta/2) - \cos(n+1/2)\theta -\sin(n+1/2)\theta)}{cos(\theta/2) - i\sin(\theta/2) - i\sin(\theta/2) -\cos(\theta/2)}

    I don't think I made any mistakes with the working above (god forbid), but what I really don't know is whether I should be taking the real/imaginary part of the expression. The real part give the answer at the numerator, but what about the denominator?

    \text{taking the real/imaginary part (?)} = \dfrac{\cos(\theta/2)-\cos(n+1/2)\theta}{-2i\sin(\theta/2)}}

    Please help. Thank you!
  12. DFranklin's Avatar
    12 14-06-2012  14:29
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    Re: Summation of trigonometric identities
    You need to take the imaginary part of the whole expression. (You can multiply top and bottom by i to make the denominator real).
  13. johnconnor92's Avatar
    13 15-06-2012  02:42
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    Re: Summation of trigonometric identities
    (Original post by DFranklin)
    You need to take the imaginary part of the whole expression. (You can multiply top and bottom by i to make the denominator real).
    What? So after the multiplication of the e^(-i theta/2) factor i STILL have to multiply another i factor into the expression? Where can I learn more about this? Thank you!

    On a side note, how does a fraction with an imaginary denominator differ from that of a real denominator? Are there any significant changes in its behaviour when the numerator/denominator changes by 1 or 2?
  14. johnconnor92's Avatar
    14 19-06-2012  03:18
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    Re: Summation of trigonometric identities
    (Original post by DFranklin)
    You need to take the imaginary part of the whole expression. (You can multiply top and bottom by i to make the denominator real).
    (Original post by johnconnor92)
    What? So after the multiplication of the e^(-i theta/2) factor i STILL have to multiply another i factor into the expression? Where can I learn more about this? Thank you!

    On a side note, how does a fraction with an imaginary denominator differ from that of a real denominator? Are there any significant changes in its behaviour when the numerator/denominator changes by 1 or 2?
    Anyone? Thanks in advance.
  15. aznkid66's Avatar
    15 19-06-2012  03:33
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    Re: Summation of trigonometric identities
    (Original post by johnconnor92)
    On a side note, how does a fraction with an imaginary denominator differ from that of a real denominator? Are there any significant changes in its behaviour when the numerator/denominator changes by 1 or 2?
    Hm? Rationalizing the denominator doesn't change the number, it's just that you cannot identify the real/imaginary part before you rationalize the denominator.

    For example, take the complex number z:

    z=\dfrac{a+bi}{c+di} Im(z)\not=\dfrac{b}{d}

    Rather:

    z=\dfrac{a+bi}{c+di}=\dfrac{ac+b d-adi+bci}{c^2+d^2} Im(z)=\dfrac{-ad+bc}{c^2+d^2}

    As always, sorry if I misunderstood your question.
  16. johnconnor92's Avatar
    16 19-06-2012  15:15
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    Re: Summation of trigonometric identities
    (Original post by aznkid66)
    Hm? Rationalizing the denominator doesn't change the number, it's just that you cannot identify the real/imaginary part before you rationalize the denominator.

    For example, take the complex number z:

    z=\dfrac{a+bi}{c+di} Im(z)\not=\dfrac{b}{d}

    Rather:

    z=\dfrac{a+bi}{c+di}=\dfrac{ac+b d-adi+bci}{c^2+d^2} Im(z)=\dfrac{-ad+bc}{c^2+d^2}

    As always, sorry if I misunderstood your question.
    That's exactly what I was confused about! DFranklin asked a factor of e^{i theta/2), and doing so produces a pure imaginary denominator which, surprisingly, gives the required answer in imaginary form. But why?
  17. DFranklin's Avatar
    17 19-06-2012  16:08
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    Re: Summation of trigonometric identities
    (Original post by johnconnor92)
    That's exactly what I was confused about! DFranklin asked a factor of e^{i theta/2), and doing so produces a pure imaginary denominator which, surprisingly, gives the required answer in imaginary form. But why?
    I'm really not sure what your problem is, which is why I haven't responded..

    If we want to find the imaginary part of \dfrac{a+ib}{id}, we can multiply top and buttom by i: \dfrac{ai-b}{-d} and then read off the imaginary part (a/-d) = -a/d.
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  18. johnconnor92's Avatar
    18 20-06-2012  07:43
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    (Original post by DFranklin)
    I'm really not sure what your problem is, which is why I haven't responded..

    If we want to find the imaginary part of \dfrac{a+ib}{id}, we can multiply top and buttom by i: \dfrac{ai-b}{-d} and then read off the imaginary part (a/-d) = -a/d.
    Oh so now i see your trick you hinted. This is amazing! Thank you so very much for teaching me this! :yes:
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