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

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- johnweldon1993

|dw:1433989500490:dw|

- 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

it would be in the same place

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## More answers

- 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

okay but what about if it wasn't 2pi? what would you do if it was something where you couldn't ignore it?

- 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

what do you mean an odd function?

- freckles

cos is an even function
sin is an odd function

- johnweldon1993

Oops, yup my mistake there!

- anonymous

okay how does that help?

- johnweldon1993

It is an identity
\[\large cos(-\theta) = cos(\theta)\]

- 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

i meant it being an even function

- 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

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

What if it was sin(-θ) = sin θ

- johnweldon1993

Ahh good question...THAT is an incorrect statement...sin(x) is actually an odd function...and I'll show you why

- 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

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

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

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

@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

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

Sorry got cut off there
supposed to be \(\large = cos(\theta)\)

- anonymous

why do the sin cos(pi/2) cancel

- johnweldon1993

What is cos(pi/2)?

- anonymous

sinθ cos(π/2)

- anonymous

there's a slash so i assume that's supposed to mean it cancels

- johnweldon1993

No I know what you were referring to :P I meant evaluate...what is cos(pi/2) ?

- anonymous

Oh haha. umm is it 0 since pi/2 is at 90 degrees or (0,1)?

- johnweldon1993

Right...and cos(90degrees) = 0
so that makes the \(\large sin(\theta)cos(\pi/2)\) cancel

- anonymous

Ahhh that makes sense

- 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

Thanks so much for all your help! :)

- johnweldon1993

Not a problem :)
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- anonymous

Thanks :)

- johnweldon1993

Didn't fully come out...but ehh :P XD

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