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# Trig Question

1. 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.
2. (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.
3. (Original post by Fatts13)
Trying to answer the exact same question.
What have you got so far?
4. (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.
5. (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.
6. (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.
7. (Original post by Reety)
Been trying to do that, but no luck so far :/
5x = 2(5x/2)
8. (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?
9. (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)
10. (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)
11. (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
12. (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.
13. (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.
14. (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
15. (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
16. (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?
17. (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?
18. (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
19. (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.
20. (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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