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March 29, 2024, 04:51:10 am

Author Topic: 3U Trig  (Read 5355 times)  Share 

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prabhleenkaur~

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3U Trig
« on: January 22, 2017, 04:47:50 pm »
+2
Hi!

Could I please get help on this question?

sin2x + sin3x + sinx =0
using general solutions

Thank you.

kiwiberry

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Re: 3U Trig
« Reply #1 on: January 22, 2017, 05:56:22 pm »
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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
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jamonwindeyer

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Re: 3U Trig
« Reply #2 on: January 22, 2017, 06:48:07 pm »
+1
Thanks kiwiberry, you are a legend!! ;D prabhleenkaur~, welcome to the forums! ;D

hanaacdr

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Re: 3U Trig
« Reply #3 on: January 23, 2017, 04:01:20 pm »
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Hi
i have a similar question,
would i be able to get some help on this,

sin2x + sin3x + sin4x = 0

much appreciated thank you!

jakesilove

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Re: 3U Trig
« Reply #4 on: January 23, 2017, 04:46:50 pm »
0
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.
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hanaacdr

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Re: 3U Trig
« Reply #5 on: January 23, 2017, 04:49:58 pm »
0
this is the question
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.

kiwiberry

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Re: 3U Trig
« Reply #6 on: January 23, 2017, 04:55:55 pm »
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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!!!
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jakesilove

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Re: 3U Trig
« Reply #7 on: January 23, 2017, 04:58:58 pm »
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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!
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ellipse

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Re: 3U Trig
« Reply #8 on: January 23, 2017, 05:03:54 pm »
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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
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ellipse

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Re: 3U Trig
« Reply #9 on: January 23, 2017, 05:04:50 pm »
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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
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jamonwindeyer

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Re: 3U Trig
« Reply #10 on: January 23, 2017, 07:26:14 pm »
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I think the above solutions are the right way to go, I can't get it to behave any simpler either :P

Shadowxo

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Re: 3U Trig
« Reply #11 on: January 23, 2017, 08:01:33 pm »
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I tried it too, I think this is just one of those questions you have to do the long way like kiwiberry did
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Re: 3U Trig
« Reply #12 on: January 23, 2017, 09:14:08 pm »
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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

hanaacdr

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Re: 3U Trig
« Reply #13 on: January 23, 2017, 10:43:00 pm »
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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

hanaacdr

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Re: 3U Trig
« Reply #14 on: January 23, 2017, 10:43:32 pm »
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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!