# Trig QuestionWatch

#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.
0
2 years ago
#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.
0
#3
(Original post by Fatts13)
Trying to answer the exact same question.
What have you got so far?
0
2 years ago
#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.
0
2 years ago
#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.
0
#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.
0
2 years ago
#7
(Original post by Reety)
Been trying to do that, but no luck so far :/
5x = 2(5x/2)
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?
0
2 years ago
#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)
0
2 years ago
#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)
0
2 years ago
#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
0
2 years ago
#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.
1
2 years ago
#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.
0
2 years ago
#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
0
2 years ago
#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
0
2 years ago
#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?
0
2 years ago
#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?
0
2 years ago
#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
0
2 years ago
#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.
0
2 years ago
#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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