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Are these equal?

  • Thread starter jtart2
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  • #1
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I'm verify some trigonometry equations and am confused about a couple of things. (This is self-study, I'm not in school)

The equation cos4x = 1+8cos^4-8cos^2 can be solved by re-writing as 2(cos2x)^2 -1 and factoring out which yields the correct answer, however based on what i've seen in other double angle identity equations one can re-write cos4x as cos(2x + 2x).

I believe this can be written as [(2cos^2 - 1)(2cos^2-1)] + [(2cos^2-1)(2cos^2-1)], however the answer then comes out to 8cos^4-8cos^2+2. So it's off by "+1".

What about my thinking is flawed? They can't both be correct!

Thanks for your help.

Joe
 

Answers and Replies

  • #2
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cos(a+b)=cosacosb-sinasinb ,what are you doing with second one in cos(2x+2x)
 
  • #3
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I'm obviously not writing the formula out correctly since the answer is not the same. I was just solving how I'v written it out [(2cos^2 - 1)(2cos^2-1)] + [(2cos^2-1)(2cos^2-1)]

Is there another way to write out cos(4x) as cos(2x+2x) instead of cos 2[(cos2x)^2-1] and have the answer come out to be 1+ 8cos^2x - 8cos^2x?

How do you add cos(2x + 2x)? or does cos4x not equal cos(2x+2x)?

Thanks,
Joe
 
Last edited:
  • #4
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This section of PF library would be of help to you. It lists all the identities you need.
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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I'm obviously not writing the formula out correctly since the answer is not the same. I was just solving how I'v written it out [(2cos^2 - 1)(2cos^2-1)] + [(2cos^2-1)(2cos^2-1)]

Is there another way to write out cos(4x) as cos(2x+2x) instead of cos 2[(cos2x)^2-1] and have the answer come out to be 1+ 8cos^2x - 8cos^2x?

How do you add cos(2x + 2x)? or does cos4x not equal cos(2x+2x)?

Thanks,
Joe
cos(2x + 2x) ≠ cos(2x) + cos(2x)

You appear to be assuming that they are equal !
 
  • #6
CAF123
Gold Member
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How do you add cos(2x + 2x)? or does cos4x not equal cos(2x+2x)?
Yes, and to expand do what user andrien suggested.
[tex] cos(2x + 2x) = cos2xcos2x -sin2xsin2x = cos^2 2x - sin^2 2x[/tex] and use known trigonometric formula to simplify the expression down to one involving only [itex] cos, [/itex] as required.
 
  • #7
Mentallic
Homework Helper
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Is there another way to write out cos(4x) as cos(2x+2x) instead of cos 2[(cos2x)^2-1]
[tex]\cos(4x)\neq \cos\left(2(\cos^2(2x)-1)\right)[/tex]

Because you already know that [itex]\cos(2x)=2cos^2(x)-1[/itex] hence [itex]2\cos^2(2x)-1=\cos(4x)[/itex] (do you see how that works?) and so finally, if we plug this expression into
[tex]\cos\left(2(\cos^2(2x)-1)\right)[/tex]
we will have that equivalent to
[tex]\cos\left(2\cos^2(2x)-2)\right)[/tex][tex]=\cos\left(2\cos^2(2x)-1+1)\right)[/tex][tex]=\cos\left(\cos(4x)+1)\right)[/tex]
 

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