Post by: prabhleenkaur~ on January 22, 2017, 04:47:50 pm
Could I please get help on this question?
sin2x + sin3x + sinx =0
using general solutions
Thank you.
Post by: kiwiberry on January 22, 2017, 05:56:22 pm
Hi!
Could I please get help on this question?
sin2x + sin3x + sinx =0
using general solutions
Thank you.
sinx + 2sinxcosx + 3sinx - 4sin3x = 0
2sinx(2 + cosx - 2sin2x) = 0
2sinx[-2(1-cos2x) + cosx + 2) = 0
sinx(2cos2x + cosx) = 0
sinxcosx(2cosx +1) = 0
therefore sinx=0, cosx=0 or cosx=-1/2
sinx = 0 = sin0
∴ x = πn
cosx = 0 = cos(π/2)
∴ x = 2πn ± π/2
cosx = -1/2 = cos(2π/3)
∴ x = 2πn ± 2π/3
Post by: jamonwindeyer on January 22, 2017, 06:48:07 pm
Post by: hanaacdr on January 23, 2017, 04:01:20 pm
i have a similar question,
would i be able to get some help on this,
sin2x + sin3x + sin4x = 0
much appreciated thank you!
Post by: jakesilove on January 23, 2017, 04:46:50 pm
Hi
i have a similar question,
would i be able to get some help on this,
sin2x + sin3x + sin4x = 0
much appreciated thank you!
For a question like this, I would expand out into just sin(x) and cos(x)
Well, this was a bad idea. I'll leave this here anyway.
Let's break it down into sin(2x)?
Nup. Did you type out the question correctly? If I wolfram alpha the solution, it's pretty damn complicated.
Post by: hanaacdr on January 23, 2017, 04:49:58 pm
i think i wrote it up correctly..
For a question like this, I would expand out into just sin(x) and cos(x)
Well, this was a bad idea. I'll leave this here anyway.
Let's break it down into sin(2x)?
Nup. Did you type out the question correctly? If I wolfram alpha the solution, it's pretty damn complicated.
Post by: kiwiberry on January 23, 2017, 04:55:55 pm
Hi
i have a similar question,
would i be able to get some help on this,
sin2x + sin3x + sin4x = 0
much appreciated thank you!
firstly, sin4x = sin(2*2x)
= 2sin2xcos2x
= 2(2sinxcosx)(2cos2x-1)
= 4sinxcosx(2cos2x-1)
sin3x = 3sinx - 4sin3x
sin2x = 2sinxcosx
let sinx=s and cosx=c
sin2x + sin3x + sin 4x
= 2sc + 3s - 4s3 + 4sc(2c2-1)
= 2sc + 3s - 4s3 + 8sc3 - 4sc
= s(3 - 2c - 4s2 + 8c3)
= s(8c3 - 2c + 4c2 - 1)
= s[2c(4c2-1) + (4c2-1)]
= s(4c2-1)(2c+1)
= s(2c-1)(2c+1)2
not sure if this is right I'm typing on my phone, someone please check!!!
Post by: jakesilove on January 23, 2017, 04:58:58 pm
firstly, sin4x = sin(2*2x)
= 2sin2xcos2x
= 2(2sinxcosx)(2cos2x-1)
= 4sinxcosx(2cos2x-1)
sin3x = 3sinx - 4sin3x
sin2x = 2sinxcosx
let sinx=s and cosx=c
sin2x + sin3x + sin 4x
= 2sc + 3s - 4s3 + 4sc(2c2-1)
= 2sc + 3s - 4s3 + 8sc3 - 4sc
= s(3 - 2c - 4s2 + 8c3)
= s(8c3 - 2c + 4c2 - 1)
= s[2c(4c2-1) + (4c2-1)]
= s(4c2-1)(2c+1)
= s(2c-1)(2c+1)2
not sure if this is right I'm typing on my phone, someone please check!!!
That's also what I get to, I just figured it was too complicated!
Post by: ellipse on January 23, 2017, 05:03:54 pm
Post by: ellipse on January 23, 2017, 05:04:50 pm
You could try using sums to products trig identities (turn sin2x+sin4x into 2sin6xcosx). Not really sure if this is 3u stuff though, although it is in the 3u hsc Cambridge book
i meant 2sin3xcosx
Post by: jamonwindeyer on January 23, 2017, 07:26:14 pm
Post by: Shadowxo on January 23, 2017, 08:01:33 pm
Post by: jamonwindeyer on January 23, 2017, 09:14:08 pm
I tried it too, I think this is just one of those questions you have to do the long way like kiwiberry did
Gross. To the OP of that question, no stress, that's never going to show up in an exam. It's just too long to be practical! ;D
Post by: hanaacdr on January 23, 2017, 10:43:00 pm
I tried it too, I think this is just one of those questions you have to do the long way like kiwiberry did
thank you! much appreciated
Post by: hanaacdr on January 23, 2017, 10:43:32 pm
Gross. To the OP of that question, no stress, that's never going to show up in an exam. It's just too long to be practical! ;D
thank you!
Post by: RuiAce on January 24, 2017, 06:10:38 pm
This sort of stuff is GENERALLY considered "Harder 3U" material for the 4U course. It is very unlikely that it should be examinable.
The more common thing to do is what ellipse said; apply a sum-to-product formula. But 3U students are, in general, without HELP, NOT expected to know the sum-to-product formulae off by heart. Only 4U students are 'expected' to memorise these procedures.
(Even then, I've only seen it appear once, which was in my trial HSC for MX2)
Post by: prabhleenkaur~ on January 27, 2017, 09:47:19 pm
sinx + 2sinxcosx + 3sinx - 4sin3x = 0
2sinx(2 + cosx - 2sin2x) = 0
2sinx[-2(1-cos2x) + cosx + 2) = 0
sinx(2cos2x + cosx) = 0
sinxcosx(2cosx +1) = 0
therefore sinx=0, cosx=0 or cosx=-1/2
sinx = 0 = sin0
∴ x = πn
cosx = 0 = cos(π/2)
∴ x = 2πn ± π/2
cosx = -1/2 = cos(2π/3)
∴ x = 2πn ± 2π/3
Thank youuu!! Appreciate it!
Post by: prabhleenkaur~ on January 27, 2017, 09:50:05 pm
Thanks kiwiberry, you are a legend!! ;D prabhleenkaur~, welcome to the forums! ;D
Glad to be a part of this ATAR Notes journey :D
Post by: armtistic on February 04, 2017, 12:28:38 pm
Hi
i have a similar question,
would i be able to get some help on this,
sin2x + sin3x + sin4x = 0
much appreciated thank you!
Rewrite it as
sin(3x-x) + sin3x + sin(3x+x) =0
Expand those and you get
2sin3xcosx+sin3x=0
sin3x(2cosx+1)=0
Solve sin3x=0 and 2cosx+1=0
x=kπ/3