Trig Question

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.

Reply 2

Reety

OP

12

Original post by Fatts13

Trying to answer the exact same question.

What have you got so far?

Reply 3

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.

Reply 4

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.

Reply 5

Reety

OP

12

Original post by 1 8 13 20 42

Express Sin5x in a different way. This will allow you to factorise.

Okay I think I've got something now.

(edited 7 years ago)

Reply 6

Original post by Reety

Been trying to do that, but no luck so far :/

5x = 2(5x/2)

Reply 7

Reety

OP

12

Original post by 1 8 13 20 42

5x = 2(5x/2)

So i got: x= +-4/5 n*pi and x= pi/6 +-2n*pi/3

Is that right?

Reply 8

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)

Reply 9

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)

Reply 10

Original post by 1 8 13 20 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 :s-smilie:

Reply 11

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 :s-smilie:

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.

Reply 12

Original post by 1 8 13 20 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.

Reply 13

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

Reply 14

Original post by 1 8 13 20 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

Reply 15

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?

Reply 16

Original post by 1 8 13 20 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?

Reply 17

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

Reply 18

Original post by 1 8 13 20 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.

Reply 19