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Simplifying a trigonometric expression

  • Thread starter Petkovsky
  • Start date
  • #1
62
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cos(4x) + [tex]\sqrt{2}[/tex]cos(3x) + cos(2x) = 0
---------------------------------------------------

Ok, so break down everything to a double angle and i get:

2cos^2(2x) - 1 + sqrt(2)*(sqrt((cos(2x) + cos^2(2x))/2) - sqrt((1-cos^(2x)*(1-cos(2x))/2) + cos(2x) = 0

...which is quite complicated to solve even if i substitute cos(2x) with 't'.

I also tried another aproach:

2cos^2(2x) - 1 + sqrt(2)*(cos(2x)cos(x) - sin(2x)(sin(x)) + 2cos(2x) - 1 = 0

here i aimed to complete a square but i cant see how.

Can you please give me some advice on how to continue or use another method
 

Answers and Replies

  • #2
238
0
Hello.

I used the Sum and Difference Formula for cosines.
Def: cos(a)+cos(b) = 2*cos((a+b)/2)*cos((a-b)/2)

Then, I let a=4x and b=2x

The original equation's LHS becomes
= 2cos((4x+2x)/2)*cos(4x-2x/2)) + [tex]\sqrt{2}[/tex]cos(3x)
= 2cos(3x)cos(2x) + [tex]\sqrt{2}[/tex]cos(3x) by simplifying
= cos(3x)[2cos(2x) + [tex]\sqrt{2}[/tex]] by distributive property
Thus, cos(3x) = 0 or 2cos(2x) + [tex]\sqrt{2}[/tex] = 0

Hope this helped.
 
Last edited:
  • #3
62
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Thank you
 
  • #4
62
0
Hello.

= 2cos((4x+2x)/2)*cos(4x-2x/2)) + [tex]\sqrt{2}[/tex]cos(3x)
= 2cos(3x)cos(2x) + [tex]\sqrt{2}[/tex]cos(3x) << -- didnt you make a mistake here?
Didnt you make a mistake?
Shouldn't it be = 2cos(3x)cos(x) + [tex]\sqrt{2}[/tex]cos(3x)?

Anyway thanx for the help.
 
  • #5
238
0
Didnt you make a mistake?
Shouldn't it be = 2cos(3x)cos(x) + [tex]\sqrt{2}[/tex]cos(3x)?

Anyway thanx for the help.
Yes, sorry.
 
  • #6
62
0
sin(3x) = cos(x-30)
How about this one?

Where should I start from?
 
  • #7
62
0
Ok i got it :)
 

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