What is the trig identity for cos^2x=(half angle)???
help!

- anonymous

What is the trig identity for cos^2x=(half angle)???
help!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- abb0t

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf

- anonymous

\[\cos(2x)\] or
\[\cos^2(x)\]?

- anonymous

@satellite73 ???

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

which one is it?

- anonymous

oh, it's the latter

- anonymous

the only thing i know to do with \(\cos^2(x)\) is to rewrite is as \(1-\sin^2(x)\)

- abb0t

http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet_Reduced.pdf

- abb0t

O.O!

- 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

it has nothing to do with double angle
it comes from the basic
\[\sin^2(x)+\cos^2(x)=1\]

- anonymous

the half angle formula comes from that?!? relly?

- anonymous

how

- anonymous

i didn't mention the half angle formula

- anonymous

the half angle formula comes from taking the double angle formula for cosine and replacing \(2x\) by \(x\)

- anonymous

...

- anonymous

?

- anonymous

You mean the cos2x=cos^2 x - sin^2 x ?

- anonymous

\[\cos(2x)=\cos^2(x)-\sin^2(x)\] replace \(2x\) by \(x\) and \(x\) therefore by \(\frac{x}{2}\)

- anonymous

then are you supposed to square it?

- anonymous

Ok I plugged them in... Now what?

- anonymous

ok hold the phone i gave you the wrong formula to use

- anonymous

\[\cos(2x)=2\cos^2(x)-1\] use that one

- anonymous

oh, ok. Will the other one not work even though they're the same?

- anonymous

no the other has both sine and cosine
no help

- anonymous

oh, ok that makes sense LOL

- anonymous

you get
\[\cos(x)=2\cos^2(\frac{x}{2})-1\] and you solve this for \(\cos(\frac{x}{2})\)

- anonymous

ok done

- anonymous

ahhh and THIS is where you need \(\pm\)!!

- anonymous

Oh! hahaha good thing I asked that before hand

- anonymous

wait... where does the plus and minus fit in?

- anonymous

I got cos^2 (x/2)=[ cos(x)+1 ]/2

- anonymous

yeah right

- anonymous

oh... the next step

- anonymous

but you want to get rid of the square
yes next step

- 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

by looking at the domain that they give you in the problem, right? What do you look at to determine that?

- anonymous

This is what I got:
cos(x/2)= (+-) ROOT[(cosx +1)/2]

- anonymous

what quadrant your angle is in

- anonymous

\[\cos(x)=\pm\sqrt{\frac{\cos(x)+1}{2}}\]

- anonymous

course it is on the cheat sheet as well

- 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

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

you lost me

- anonymous

yeah, it's what I do with my sin^2 +cos^2=1

- anonymous

ok wait a sec

- anonymous

cos^2 x=(1/2) (1+cos(2x))

- anonymous

thats wat it says on the sheet

- anonymous

fine
then take the square root and get exactly what you got before

- anonymous

oh, and replace \(x\) by \(\frac{x}{2}\)

- anonymous

oh! So THAT's what happened! LOL that was simple.

- anonymous

I was like, why is there a double angle in this formula..? hahaha

- anonymous

yeah simple enough

- anonymous

why trig so early in the semester?

- 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

have fun

- anonymous

Are you leaving?

- anonymous

:) thanks for your time~ I really appreciate it a lot!

Looking for something else?

Not the answer you are looking for? Search for more explanations.