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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433989500490:dw

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2The unit circle Let me ask...if you start dw:1433989606182:dw And you travel 2π around the circle...where will you end up?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it would be in the same place

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Right...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
 one year ago
Best ResponseYou've already chosen the best response.0okay but what about if it wasn't 2pi? what would you do if it was something where you couldn't ignore it?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean an odd function?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1cos is an even function sin is an odd function

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Oops, yup my mistake there!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay how does that help?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2It is an identity \[\large cos(\theta) = cos(\theta)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0i meant it being an even function

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2So 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
 one year ago
Best ResponseYou've already chosen the best response.0What if it was sin(θ) = sin θ

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Ahh good question...THAT is an incorrect statement...sin(x) is actually an odd function...and I'll show you why

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2dw: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
 one year ago
Best ResponseYou've already chosen the best response.2SOOOO 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
 one year ago
Best ResponseYou've already chosen the best response.2Which 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
 one year ago
Best ResponseYou've already chosen the best response.0Okay. 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
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.2No 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
 one year ago
Best ResponseYou've already chosen the best response.2Sorry got cut off there supposed to be \(\large = cos(\theta)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do the sin cos(pi/2) cancel

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2What is cos(pi/2)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there's a slash so i assume that's supposed to mean it cancels

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2No I know what you were referring to :P I meant evaluate...what is cos(pi/2) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh haha. umm is it 0 since pi/2 is at 90 degrees or (0,1)?

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Right...and cos(90degrees) = 0 so that makes the \(\large sin(\theta)cos(\pi/2)\) cancel

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh that makes sense

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Good :) So as long as you can remember those sum and difference formulas for these types of problems...they will go by pretty quick!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much for all your help! :)

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Not a problem :) \[\bbox [10pt, black, border:5pt solid blue]{\Huge\cal\color{red}{\bigstar}\color{orange}{G}\color{yellow}{O}\color{green}{O}\color{blue}{D} \space \color{indigo}{L}\color{violet}{U}\color{red}{C}\color{orange}{K}\space \color{yellow}{T}\color{green}{O}\color{blue}{M}\color{indigo}{O}\color{violet}{R}\color{red}{R}\color{orange}{O}\color{yellow}{W}\color{blue}{\bigstar}}\]

johnweldon1993
 one year ago
Best ResponseYou've already chosen the best response.2Didn't fully come out...but ehh :P XD
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