Reexpress 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.
Trig Question
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 3 weeks ago 3w ago

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 3 weeks ago 3w ago
(Original post by Reety)
Reexpress 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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 3 weeks ago 3w ago
(Original post by Fatts13)
Trying to answer the exact same question. 
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 4
 3 weeks ago 3w ago
(Original post by Reety)
Reexpress 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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 5
 3 weeks ago 3w ago
(Original post by Reety)
What have you got so far?
sin2theta + sin3theta = 2sin(5/2theta)cos(1/2theta)
Dont know where to go from there. 
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 3 weeks ago 3w ago
(Original post by 13 1 20 8 42)
Express Sin5x in a different way. This will allow you to factorise.Last edited by Reety; 3 weeks ago at 15:34. 
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 3 weeks ago 3w ago
(Original post by Reety)
Been trying to do that, but no luck so far :/Post rating:2 
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 3 weeks ago 3w ago
(Original post by 13 1 20 8 42)
5x = 2(5x/2)
Is that right? 
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 3 weeks ago 3w ago
(Original post by Fatts13)
I did:
sin2theta + sin3theta = 2sin(5/2theta)cos(1/2theta)
Dont know where to go from there. 
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 3 weeks ago 3w ago

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 3 weeks ago 3w ago
(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)
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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 3 weeks ago 3w ago
(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
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.Post rating:1 
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 13
 3 weeks ago 3w ago
(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.
I haven't done maths for over a year  my maths is really rusty. 
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 14
 3 weeks ago 3w ago
(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. 
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 3 weeks ago 3w ago
(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
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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 16
 3 weeks ago 3w ago
(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
I'm not sure how you got that solution. Did you do the same rewriting process as before? 
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 3 weeks ago 3w ago
(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?
For 2sin(5x/2) = 0 the other solution should be x=2pi/5 + 4pin/5? No? 
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 3 weeks ago 3w ago
(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?
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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 3 weeks ago 3w ago
(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
Sorry again. 
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 3 weeks ago 3w ago
(Original post by Fatts13)
Ohh OK I get it now. So the solutions are: 4npi/5, 2npi5 and =pi/6 +2npi/3 ?
Sorry again.
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Updated: October 3, 2016
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