Solving Trigonometry Equations: Confused about Cos4x?

[jtart2]

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

 

[andrien]

cos(a+b)=cosacosb-sinasinb ,what are you doing with second one in cos(2x+2x)

 

[jtart2]

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

 

[Saitama]

This section of PF library would be of help to you. It lists all the identities you need.

 

[SammyS]

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 !

 

[CAF123]

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

Yes, and to expand do what user andrien suggested.
cos(2x + 2x) = cos2xcos2x -sin2xsin2x = cos^2 2x - sin^2 2x and use known trigonometric formula to simplify the expression down to one involving only cos, as required.

 

[Mentallic]

Is there another way to write out cos(4x) as cos(2x+2x) instead of cos 2[(cos2x)^2-1]

\cos(4x)\neq \cos\left(2(\cos^2(2x)-1)\right)

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