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Math Help - Solve sinx+ sin2x+ sin3x= 1

  1. #1
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    Solve sinx+ sin2x+ sin3x= 1

    sinx+ sin2x+ sin3x= 1
    thanks for all
    Last edited by mr fantastic; July 24th 2011 at 07:02 PM. Reason: Re-titled.
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  2. #2
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    Re: please help me this trigonometric

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

    \displaystyle \sin 3x = 3\sin x - 4\sin^3 x
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    Re: please help me this trigonometric

    can you help me complete? it not so easy like that
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: please help me this trigonometric

    Quote Originally Posted by quynhnguyen View Post
    sinx+ sin2x+ sin3x= 1
    thanks for all
    You can also prove by induction that:


    \sin(x)+\sin(2x)+\sin(3x)+...+\sin(nx)=\frac{\sin{  (\frac{n}{2}x)}\sin(\frac{n+1}{2}x)}{\sin(\frac{x}  {2})}
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    Re: please help me this trigonometric

    please help me complete, : (
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    Re: please help me this trigonometric

    Quote Originally Posted by quynhnguyen View Post
    please help me complete, : (
    Post #2 puts you on the right track. I'll add that line 2 of that reply follows from using the compound angle formula on sin(x + 2x).

    You need to show some effort if more help is needed. What have you tried and where are you still stuck? Are you expected to solve it by hand or using CAS technology?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: please help me this trigonometric

    Hint!

    I will post the whole solution soon, but take a moment and think why multiplying the both sides of an equation with 2\sin{\frac{x}{2}} will help you...
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: please help me this trigonometric

    Not the whole solution...

    With help of my formula we get:

    \sin2x\sin{\frac{3x}{2}}=\sin{\frac{x}{2}}

    2\sin{x}\cos{x}\sin{\frac{3x}{2}}=\sin{\frac{x}{2}  }

    4\sin {\frac{x}{2}} \cos{\frac{x}{2}} \cos{x} \sin {\frac{3x}{2}}=\sin{\frac{x}{2}}



    \sin {\frac{x}{2}} (4\cos {\frac{x}{2}} \cos {x} \sin {\frac{3x}{2}}-1)=0


    \sin {\frac{x}{2}} (4\cos {\frac{x}{2}}\cos {x} \sin {\frac{3x}{2}}-1)=0

    \sin {\frac{x}{2}} (2\sin {2x}+\sin {2x}-1)=0

    \sin {\frac{x}{2}} (2 \sin {2x} \cos {x}+\sin {2x}-1)=0
    Last edited by Also sprach Zarathustra; July 24th 2011 at 08:02 PM.
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    Re: please help me this trigonometric

    Quote Originally Posted by Also sprach Zarathustra View Post
    Hint!

    I will post the whole solution soon, but take a moment and think why multiplying the both sides of an equation with 2\sin{\frac{x}{2}} will help you...
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: please help me this trigonometric

    Are you sure there's no typo? Maybe the equation is:
    \sin(x)+\sin(2x)+\sin(3x)=0
    ?
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    Re: please help me this trigonometric

    you can use the identity
    \sin{P}+\sin{Q}=2\sin(\frac{P+Q}{2})\cos(\frac{P-Q}{2})
    using arrangement \sin{3x}+\sin{x}+\sin{2x}
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    Re: please help me this trigonometric

    Quote Originally Posted by waqarhaider View Post
    you can use the identity
    \sin{P}+\sin{Q}=2\sin(\frac{P+Q}{2})\cos(\frac{P-Q}{2})
    using arrangement \sin{3x}+\sin{x}+\sin{2x}
    thanks but it not so easy like that
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  13. #13
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    Re: please help me this trigonometric

    Quote Originally Posted by Siron View Post
    Are you sure there's no typo? Maybe the equation is:
    \sin(x)+\sin(2x)+\sin(3x)=0
    ?
    yes i'm sure: sinx+ sin2x+ sin3x= 1
    if: sinx+ sin2x+ sin3x= 0 it too easy thats right? so i needn't post this trigonometri in here
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  14. #14
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    Re: please help me this trigonometric

    Quote Originally Posted by quynhnguyen View Post
    sinx+ sin2x+ sin3x= 1
    thanks for all
    This equation does not seem to have a clean solution. If you graph the function, you can see that there are two solutions between 0 and 90 (and no others in the range 0 to 360). The smaller solution is approximately 9.836, but that does not look like a recognisable quantity in either degrees or radians. I haven't tried to locate the other solution accurately, but it is somewhere near 71.3. Again, this is not a nice-looking angle. The solutions for \sin x are not rational. I doubt whether the equation can be solved exactly.
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    Re: please help me this trigonometric

    Quote Originally Posted by Opalg View Post
    This equation does not seem to have a clean solution. If you graph the function, you can see that there are two solutions between 0 and 90 (and no others in the range 0 to 360). The smaller solution is approximately 9.836, but that does not look like a recognisable quantity in either degrees or radians. I haven't tried to locate the other solution accurately, but it is somewhere near 71.3. Again, this is not a nice-looking angle. The solutions for \sin x are not rational. I doubt whether the equation can be solved exactly.
    yes . Here: \ sinx + sin2x + sin3x = 1 graph

    Solve sinx+ sin2x+ sin3x= 1-dt.jpg
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