2003 Step I Q3 Watch

Square
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STEP I June 03:

last part:

show that \displaystyle 2\cos(\frac{1}{2}\theta)=\cos(\t  heta) if and only if \displaystyle \theta =(4n+2)\pi \pm 2\rho where \cos\rho=\frac{1}{2}(\sqrt{3}-1)

\displaystyle 0\le\rho\le\frac{\pi}{2} and n any integer.

Basically, I got the inital expression down to a quadratic in cos(1/2 theta) and solved that and got solutions: \frac{-1\pm\sqrt{3}}{2} (which looks fairly similar to cos(rho)). I've puzzled it over and resisted the temptation to just look at the solutions, but I'm still stumped.

Trig never was my best area, good chance I've probably overlooked something trivial.

Either way, hints only please!
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DFranklin
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Hint: you're very close to the answer (and as far as I can see, haven't made any mistakes, so you just need to do more, you don't need to change anything).

If you want more of a hint, can you explain exactly what your sticking point is?
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Square
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(Original post by DFranklin)
Hint: you're very close to the answer (and as far as I can see, haven't made any mistakes, so you just need to do more, you don't need to change anything).

If you want more of a hint, can you explain exactly what your sticking point is?
Blargh, well I just don't know how to move on.

I've tried arccosing things but I don't see quite how that will help.

EDIT2:Another penny dropping moment: \frac{-1-\sqrt{3}}{2}<-1 which means no solutions for theta.

Im jumping the gun, scrap EDIT 1, but I still stick by EDIT 2: which surely writes off one of the solutions from the quadratic.
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Square
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Obvious steps that I can't believe I overlooked before have lead me to:

\theta=4n\pi \pm 2\rho

Which isnt quite right surely?
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Winter
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I don't know how you made the mistake.
But cos(1/2 theta) is (1-sqrt(3))/2 not (sqrt(3)-1)/2
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Square
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*facepalm*

I don't even want to write what I did, it's far too embarassing. You can probably guess anyway.
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Winter
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2cos(1/2 theta) = 2 cos(theta)^2 - 1
so 2x^2-2x-1=0 (x=cos(1/2 theta))
then x= ( 1 +/- sqrt(3) )/2
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Square
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Yep, I got it now I think, thanks.

Elementary mistake, time to write trig identities on my bedroom wall I reckon.
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DFranklin
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Sorry - looks like you did make a mistake; I thought you were closer to the right answer than you were.
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