anonymous
  • anonymous
Does... 1. sin(2pi + Ѳ) = -cos Ѳ 2. cos(2pi - Ѳ) = cos Ѳ I don't know what to do with the pi's. My teacher never taught me how to check if equations are identities with pi in them. Please explain.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
johnweldon1993
  • johnweldon1993
|dw:1433989500490:dw|
johnweldon1993
  • johnweldon1993
The unit circle Let me ask...if you start |dw:1433989606182:dw| And you travel 2π around the circle...where will you end up?
anonymous
  • anonymous
it would be in the same place

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johnweldon1993
  • johnweldon1993
Right...you would end up in the same exact place, meaning we can just ignore them and just worry about 1) \(\large sin(\theta) = -cos(\theta)\) and clearly that is NOT a correct statement Does that make sense?
anonymous
  • anonymous
okay but what about if it wasn't 2pi? what would you do if it was something where you couldn't ignore it?
johnweldon1993
  • johnweldon1993
It would be the same for 2) as well \(\large cos(2\pi - \theta) = cos(\theta)\) since 2π brings us back to the same place, we ignore it and just have \[\large cos(-\theta) = cos(\theta)\] which *since we know cos is an odd function is correct
anonymous
  • anonymous
what do you mean an odd function?
freckles
  • freckles
cos is an even function sin is an odd function
johnweldon1993
  • johnweldon1993
Oops, yup my mistake there!
anonymous
  • anonymous
okay how does that help?
johnweldon1993
  • johnweldon1993
It is an identity \[\large cos(-\theta) = cos(\theta)\]
anonymous
  • anonymous
You said since cos in an odd function it is correct? what do you mean by that? How would it being an odd function help
anonymous
  • anonymous
i meant it being an even function
johnweldon1993
  • johnweldon1993
Yeah I did too, still sorry about that typo lol So here maybe a visual would help with this...the graph of cos(x) is |dw:1433991380473:dw|
johnweldon1993
  • johnweldon1993
So for example if you have cos(π) and cos(-π) |dw:1433991571607:dw| So it really helps up simplify things like \[\large cos(-\theta) = cos(\theta)\]
anonymous
  • anonymous
What if it was sin(-θ) = sin θ
johnweldon1993
  • johnweldon1993
Ahh good question...THAT is an incorrect statement...sin(x) is actually an odd function...and I'll show you why
johnweldon1993
  • johnweldon1993
|dw:1433991809831:dw| It has symmetry about the origin...meaning if you were to rotate the right half of the graph about the origin you would end up with the left half
johnweldon1993
  • johnweldon1993
SOOOO now lets say you choose x = pi/2 for example If you go to pi/2 and -pi/2 on this graph you will get |dw:1433991965337:dw|
johnweldon1993
  • johnweldon1993
Which leads to another good identity that \[\large sin(-\theta) = -sin(\theta)\] Like here we chose theta to me pi/2 \[\large sin(-\pi/2) = -sin(\pi/2) \rightarrow -1 = -1 \checkmark \]
anonymous
  • anonymous
Okay. How would you verify something like cos (θ + (π/2)) = -sin θ. I get it with the 2π because it cancels but what about this?
anonymous
  • anonymous
@johnweldon1993 sorry that I am asking so many questions. I really want to get this stuff down because I have my final exam tomorrow
johnweldon1993
  • johnweldon1993
No that's okay...actually everything we've been doing has been just from the fact that we knew the 2π would go away...so it's good to know the others as well! EVERYTHING we have done so far and can do...is from these next things Remember those identities that say \[\large sin(a \pm b) = sin(a)cos(b) + sin(b)cos(a)\] and \[\large cos(a \pm \ b) = cos(a)cos(b) \mp sin(a)sin(b)\] Well if you have sin(theta + π/2) that would be \[\large sin(\theta + \pi /2) = \cancel{sin(\theta)cos(\pi/2)} + sin(\pi/2)cos(\theta) = (1)(cos(\theta)) = cos(\theta)\]
johnweldon1993
  • johnweldon1993
Sorry got cut off there supposed to be \(\large = cos(\theta)\)
anonymous
  • anonymous
why do the sin cos(pi/2) cancel
johnweldon1993
  • johnweldon1993
What is cos(pi/2)?
anonymous
  • anonymous
sinθ cos(π/2)
anonymous
  • anonymous
there's a slash so i assume that's supposed to mean it cancels
johnweldon1993
  • johnweldon1993
No I know what you were referring to :P I meant evaluate...what is cos(pi/2) ?
anonymous
  • anonymous
Oh haha. umm is it 0 since pi/2 is at 90 degrees or (0,1)?
johnweldon1993
  • johnweldon1993
Right...and cos(90degrees) = 0 so that makes the \(\large sin(\theta)cos(\pi/2)\) cancel
anonymous
  • anonymous
Ahhh that makes sense
johnweldon1993
  • johnweldon1993
Good :) So as long as you can remember those sum and difference formulas for these types of problems...they will go by pretty quick!
anonymous
  • anonymous
Thanks so much for all your help! :)
johnweldon1993
  • johnweldon1993
Not a problem :) \[\bbox [10pt, black, border:5pt solid blue]{\Huge\cal\color{red}{\bigstar}\color{orange}{G}\color{yellow}{O}\color{green}{O}\color{blue}{D} \space \color{indigo}{L}\color{violet}{U}\color{red}{C}\color{orange}{K}\space \color{yellow}{T}\color{green}{O}\color{blue}{M}\color{indigo}{O}\color{violet}{R}\color{red}{R}\color{orange}{O}\color{yellow}{W}\color{blue}{\bigstar}}\]
anonymous
  • anonymous
Thanks :)
johnweldon1993
  • johnweldon1993
Didn't fully come out...but ehh :P XD

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