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January 19th, 2013, 04:03 PM   #1
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Confusing Pre-Calculus Problem. Can Anyone Counter This?

Question: Sin 4x = -2 Sin 2x

What I did:
= 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula
= 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula
=2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2
= (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx
=2-2sin^2 x = 0 Added 1 on both sides.
=2(1-sinx)(1+sinx) = 0 Factored
1- sinx = 0 & sinx + 1 =0 Obvious method
sinx = 1 & sinx = -1 "
x= pi/2 U x = 3pi/2 Unit circle.


But however, the answer key says this is wrong.
Others' solutions:
Sin 4x = -2 sin 2x
= 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula
= 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x
2sin 2x = 0 & cos2x = -1
sin 2x = 0 & cos 2x = -1
0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2

x= 0, pi, pi/2, 3pi/2




Does anyone know what I did wrong? ( the first solution)
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January 19th, 2013, 04:23 PM   #2
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Re: Confusing Pre-Calculus Problem. Can Anyone Counter This?

Quote:
Originally Posted by bboyinmartin
Question: Sin 4x = -2 Sin 2x

What I did:
= 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula
Yes, that is correct. Although, personally, I wouldn't have use the "double angle formula" on the right. I would have stopped with a formula in "2x" on both sides.

Quote:
= 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula
cos(2x)= 1+ 2 cos^2(x))
=2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2
= (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx
IF . Otherwise you cannot do that. You need to check those cases, sin(x)= 0 or cos(x) 0, separately.

Quote:
[tex]=2-2sin^2 x = 0 Added 1 on both sides.
=2(1-sinx)(1+sinx) = 0 Factored
1- sinx = 0 & sinx + 1 =0 Obvious method
sinx = 1 & sinx = -1 "
x= pi/2 U x = 3pi/2 Unit circle.


But however, the answer key says this is wrong.
Others' solutions:
Sin 4x = -2 sin 2x
= 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula
= 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x
2sin 2x = 0 & cos2x = -1
sin 2x = 0 & cos 2x = -1
0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2

x= 0, pi, pi/2, 3pi/2
All cases in which either sin(x)= 0 or cos(x)= 0, and therefore sin(x)cos(x)= 0.


Quote:

Does anyone know what I did wrong? ( the first solution)
You can't divide by 0!
HallsofIvy is offline  
January 19th, 2013, 06:33 PM   #3
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Re: Confusing Pre-Calculus Problem. Can Anyone Counter This?

Quote:
Originally Posted by HallsofIvy
Quote:
Originally Posted by bboyinmartin
Question: Sin 4x = -2 Sin 2x

What I did:
= 2 (sin2x)(cos2x) = -2(2sinx cosx) Double angle formula
Yes, that is correct. Although, personally, I wouldn't have use the "double angle formula" on the right. I would have stopped with a formula in "2x" on both sides.

Quote:
= 2(2sinx cosx) (1-2sin^2x) =-2(2 sinx cosx) Double angle formula
cos(2x)= 1+ 2 cos^2(x))
=2sinx cosx (1- 2sin^2x) = -2(sinx cosx) Divided everything by 2
= (1-2sin^2x) = -1 Divided both sides by 2 sinx cosx
IF . Otherwise you cannot do that. You need to check those cases, sin(x)= 0 or cos(x) 0, separately.

[quote:2rd9xf30][tex]=2-2sin^2 x = 0 Added 1 on both sides.
=2(1-sinx)(1+sinx) = 0 Factored
1- sinx = 0 & sinx + 1 =0 Obvious method
sinx = 1 & sinx = -1 "
x= pi/2 U x = 3pi/2 Unit circle.


But however, the answer key says this is wrong.
Others' solutions:
Sin 4x = -2 sin 2x
= 2(sin2x cos2x) + 2(sin 2x) = 0 Double angle formula
= 2 sin 2x ( cos2x + 1) =0 factored our 2 sin 2x
2sin 2x = 0 & cos2x = -1
sin 2x = 0 & cos 2x = -1
0, pi, pi/2, 3pi/2 U pi/2, pi, 3pi/2

x= 0, pi, pi/2, 3pi/2
All cases in which either sin(x)= 0 or cos(x)= 0, and therefore sin(x)cos(x)= 0.


Quote:

Does anyone know what I did wrong? ( the first solution)
You can't divide by 0![/quote:2rd9xf30]





Oh, okay. Thanks!
bboyinmartin is offline  
January 19th, 2013, 06:48 PM   #4
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Math Focus: Calculus/ODEs
Re: Confusing Pre-Calculus Problem. Can Anyone Counter This?

I would use the strategy suggested by [color=#0000FF]HallsofIvy[/color]:



Use the double-angle identity for sine on the left, and arrange as:



Factor:



We have two cases to consider (where ):

i)

ii)

However, we see that all of the solutions from the second case are part of the first, so we may simply state:

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