anonymous
  • anonymous
What is the trig identity for cos^2x=(half angle)??? help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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abb0t
  • abb0t
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf
anonymous
  • anonymous
\[\cos(2x)\] or \[\cos^2(x)\]?
anonymous
  • anonymous
@satellite73 ???

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anonymous
  • anonymous
which one is it?
anonymous
  • anonymous
oh, it's the latter
anonymous
  • anonymous
the only thing i know to do with \(\cos^2(x)\) is to rewrite is as \(1-\sin^2(x)\)
abb0t
  • abb0t
http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf
abb0t
  • abb0t
O.O!
anonymous
  • anonymous
I saw the formula, but I don't know how you would derive that... and especially when you also have a double angle thrown in there... LOL
anonymous
  • anonymous
it has nothing to do with double angle it comes from the basic \[\sin^2(x)+\cos^2(x)=1\]
anonymous
  • anonymous
the half angle formula comes from that?!? relly?
anonymous
  • anonymous
how
anonymous
  • anonymous
i didn't mention the half angle formula
anonymous
  • anonymous
the half angle formula comes from taking the double angle formula for cosine and replacing \(2x\) by \(x\)
anonymous
  • anonymous
...
anonymous
  • anonymous
?
anonymous
  • anonymous
You mean the cos2x=cos^2 x - sin^2 x ?
anonymous
  • anonymous
\[\cos(2x)=\cos^2(x)-\sin^2(x)\] replace \(2x\) by \(x\) and \(x\) therefore by \(\frac{x}{2}\)
anonymous
  • anonymous
then are you supposed to square it?
anonymous
  • anonymous
Ok I plugged them in... Now what?
anonymous
  • anonymous
ok hold the phone i gave you the wrong formula to use
anonymous
  • anonymous
\[\cos(2x)=2\cos^2(x)-1\] use that one
anonymous
  • anonymous
oh, ok. Will the other one not work even though they're the same?
anonymous
  • anonymous
no the other has both sine and cosine no help
anonymous
  • anonymous
oh, ok that makes sense LOL
anonymous
  • anonymous
you get \[\cos(x)=2\cos^2(\frac{x}{2})-1\] and you solve this for \(\cos(\frac{x}{2})\)
anonymous
  • anonymous
ok done
anonymous
  • anonymous
ahhh and THIS is where you need \(\pm\)!!
anonymous
  • anonymous
Oh! hahaha good thing I asked that before hand
anonymous
  • anonymous
wait... where does the plus and minus fit in?
anonymous
  • anonymous
I got cos^2 (x/2)=[ cos(x)+1 ]/2
anonymous
  • anonymous
yeah right
anonymous
  • anonymous
oh... the next step
anonymous
  • anonymous
but you want to get rid of the square yes next step
anonymous
  • anonymous
but be careful even though there is a \(\pm\) in the formula, doesn't means you get two answers you have to decide if it is plus or minus
anonymous
  • anonymous
by looking at the domain that they give you in the problem, right? What do you look at to determine that?
anonymous
  • anonymous
This is what I got: cos(x/2)= (+-) ROOT[(cosx +1)/2]
anonymous
  • anonymous
what quadrant your angle is in
anonymous
  • anonymous
\[\cos(x)=\pm\sqrt{\frac{\cos(x)+1}{2}}\]
anonymous
  • anonymous
course it is on the cheat sheet as well
anonymous
  • anonymous
actually it is really good to know how to derive these because who can memorize all this drek easy to know a few and then get the ones you need
anonymous
  • anonymous
yes, but why does that one have a double angle in it? even if you take the root, there's a double angle in it...
anonymous
  • anonymous
you lost me
anonymous
  • anonymous
yeah, it's what I do with my sin^2 +cos^2=1
anonymous
  • anonymous
ok wait a sec
anonymous
  • anonymous
cos^2 x=(1/2) (1+cos(2x))
anonymous
  • anonymous
thats wat it says on the sheet
anonymous
  • anonymous
fine then take the square root and get exactly what you got before
anonymous
  • anonymous
oh, and replace \(x\) by \(\frac{x}{2}\)
anonymous
  • anonymous
oh! So THAT's what happened! LOL that was simple.
anonymous
  • anonymous
I was like, why is there a double angle in this formula..? hahaha
anonymous
  • anonymous
yeah simple enough
anonymous
  • anonymous
why trig so early in the semester?
anonymous
  • anonymous
Oh, and I need some geomeetry brushing up to do, I'll open a new question for it... I can memorize these geo. volume and are formulas all I want, but I won't understand thme and be able to use them...
anonymous
  • anonymous
have fun
anonymous
  • anonymous
Are you leaving?
anonymous
  • anonymous
:) thanks for your time~ I really appreciate it a lot!

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