Trig Question

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    Re-express Sin2x + Sin3x in terms of a product of trigonometric functions, and use this to find the general solution of: Sin2x + Sin3x + Sin5x = 0

    So far I've got: 2Sin(5x/2)Cos(x/2) + Sin5x = 0

    I have no clue what to do next, if anyone could help would be great.

    Thanks.
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    (Original post by Reety)
    Re-express Sin2x + Sin3x in terms of a product of trigonometric functions, and use this to find the general solution of: Sin2x + Sin3x + Sin5x = 0

    So far I've got: 2Sin(5x/2)Cos(x/2) + Sin5x = 0

    I have no clue what to do next, if anyone could help would be great.

    Thanks.
    Trying to answer the exact same question.
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    (Original post by Fatts13)
    Trying to answer the exact same question.
    What have you got so far?
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    (Original post by Reety)
    Re-express Sin2x + Sin3x in terms of a product of trigonometric functions, and use this to find the general solution of: Sin2x + Sin3x + Sin5x = 0

    So far I've got: 2Sin(5x/2)Cos(x/2) + Sin5x = 0

    I have no clue what to do next, if anyone could help would be great.

    Thanks.
    Express Sin5x in a different way. This will allow you to factorise.
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    (Original post by Reety)
    What have you got so far?
    I did:
    sin2theta + sin3theta = 2sin(5/2theta)cos(-1/2theta)
    Dont know where to go from there.
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    (Original post by 13 1 20 8 42)
    Express Sin5x in a different way. This will allow you to factorise.
    Okay I think I've got something now.
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    (Original post by Reety)
    Been trying to do that, but no luck so far :/
    5x = 2(5x/2)
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    (Original post by 13 1 20 8 42)
    5x = 2(5x/2)
    So i got: x= +-4/5 n*pi and x= pi/6 +-2n*pi/3

    Is that right?
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    (Original post by Fatts13)
    I did:
    sin2theta + sin3theta = 2sin(5/2theta)cos(-1/2theta)
    Dont know where to go from there.
    It shouldn't matter too much, but it's sort of neater to use 1/2 theta rather than -1/2 theta (of course both are correct)
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    (Original post by Reety)
    So i got: x= +-4/5 n*pi and x= pi/6 +-2n*pi/3

    Is that right?
    I got different solutions. I checked each with some values of n. The first solution you have there is indeed a correct set, I believe, but I've found x = +-2/5 n * pi seems to work. The second set doesn't seem to be correct, e.g. pi/6 doesn't work. I have three seemingly distinct sets/functions of a natural number that seem to work in total. Presumably you factorised out sin(5x/2), and sin(5x/2) = 0 clearly implies 5x/2 = n*pi for some n (note that if n is considered to be any integer the +/- is not needed) so x = 2n*pi/5. The other factor is harder to deal with and I'd recommend also turning that into a product (presumably it can be done more by "feel"/looking at the graph carefully, but screw that)
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    (Original post by 13 1 20 8 42)
    It shouldn't matter too much, but it's sort of neater to use 1/2 theta rather than -1/2 theta (of course both are correct)
    At the moment I have this:
    sin2x + sin3x +sin5x = 0
    2sin(5x/2)cos(x/2) + 2sin(5x/2) = 0
    2sin(5x/2) (cos(x/2) +1) = 0
    so 2sin(5x/2) = 0 and cos(x/2 + 1) = 0
    from there I got
    x= +- 4pin/5 and x= 2pi + 4pin
    I have no clue if that's correct
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    (Original post by Fatts13)
    At the moment I have this:
    sin2x + sin3x +sin5x = 0
    2sin(5x/2)cos(x/2) + 2sin(5x/2) = 0
    2sin(5x/2) (cos(x/2) +1) = 0
    so 2sin(5x/2) = 0 and cos(x/2 + 1) = 0
    from there I got
    x= +- 4pin/5 and x= 2pi + 4pin
    I have no clue if that's correct
    Your alternative expression for sin(5x) is wrong. Recall that Sin(2A) = 2sinAcosA.
    The first solution set is technically correct, but you can do better. Note that siny = 0 whenever y = +- npi. Presumably you used +- 2npi which is correct but incomplete.
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    (Original post by 13 1 20 8 42)
    Your alternative expression for sin(5x) is wrong. Recall that Sin(2A) = 2sinAcosA.
    The first solution set is technically correct, but you can do better. Note that siny = 0 whenever y = +- npi. Presumably you used +- 2npi which is correct but incomplete.
    So would sin(5x) be 5sinxcosx? Or is it sin(3x+2x) =sin3xcos2x + cos3xsin2x? Or just something completely different.
    I haven't done maths for over a year - my maths is really rusty.
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    (Original post by Fatts13)
    So would sin(5x) be 5sinxcosx? Or is it sin(3x+2x) =sin3xcos2x + cos3xsin2x? Or just something completely different.
    I haven't done maths for over a year - my maths is really rusty.
    No the point is that you want to use the 5x/2. Sin(5x) = sin(2(5x/2)) = 2sin(5x/2)cos(5x/2) by the formula I gave before
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    (Original post by 13 1 20 8 42)
    No the point is that you want to use the 5x/2. Sin(5x) = sin(2(5x/2)) = 2sin(5x/2)cos(5x/2) by the formula I gave before
    So sorry, I'm being a pest :/
    Again I think i got it wrong:
    sin2x + sin3x +sin5x = 02sin(5x/2)cos(x/2) + 2sin(5x/2)cos(5x/2) = 0
    2sin(5x/2) (cosx/2 + cos5x/2) = 0
    2sin(5x/2) = 0 => x = 4pin/5
    cos(x/2) + cos(5x/2) = 0 => cos3x = 0 => x= pi/6 + 2pi/3
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    (Original post by Fatts13)
    So sorry, I'm being a pest :/
    Again I think i got it wrong:
    sin2x + sin3x +sin5x = 02sin(5x/2)cos(x/2) + 2sin(5x/2)cos(5x/2) = 0
    2sin(5x/2) (cosx/2 + cos5x/2) = 0
    2sin(5x/2) = 0 => x = 4pin/5
    cos(x/2) + cos(5x/2) = 0 => cos3x = 0 => x= pi/6 + 2pi/3
    As said before, you seem to be missing some solutions from the 2sin(5x/2) = 0. Recall that siny = 0 if y = pi*n.
    I'm not sure how you got that solution. Did you do the same rewriting process as before?
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    (Original post by 13 1 20 8 42)
    As said before, you seem to be missing some solutions from the 2sin(5x/2) = 0. Recall that siny = 0 if y = pi*n.
    I'm not sure how you got that solution. Did you do the same rewriting process as before?
    In all honesty I have no idea what i did.
    For 2sin(5x/2) = 0 the other solution should be x=2pi/5 + 4pin/5? No?
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    (Original post by Fatts13)
    In all honesty I have no idea what i did.
    For 2sin(5x/2) = 0 the other solution should be x=2pi/5 + 4pin/5? No?
    Just turn the sum of the cosines into a single product in exactly the same way as you turned Sin2x + Sin3x into a single product.

    2sin(5x/2) = 0 -> sin(5x/2) = 0
    Sin(5x/2) = 0 <-> 5x/2 = n*pi (think of a graph of sinx, it meets the x axis at n*pi for all integer values n)
    --> x = 2n*pi/5
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    (Original post by 13 1 20 8 42)
    Just turn the sum of the cosines into a single product in exactly the same way as you turned Sin2x + Sin3x into a single product.

    2sin(5x/2) = 0 -> sin(5x/2) = 0
    Sin(5x/2) = 0 <-> 5x/2 = n*pi (think of a graph of sinx, it meets the x axis at n*pi for all integer values n)
    --> x = 2n*pi/5
    Ohh OK I get it now. So the solutions are: 4npi/5, 2npi5 and =-pi/6 +2npi/3 ?
    Sorry again.
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    (Original post by Fatts13)
    Ohh OK I get it now. So the solutions are: 4npi/5, 2npi5 and =-pi/6 +2npi/3 ?
    Sorry again.
    The first solution set is redundant because all such solutions are contained in the second. The last is incorrect. pi/6 does not work, for instance. For the other solutions you should get cos(x/2) + cos(5x/2) = 0 so 2 * cosx * cos(3x/2) = 0.
 
 
 
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