cot(-8pi) find the exact value
Do you know what the cosine of \(8\pi\) is?
Nope! Honestly, I have no clue what's happening in class.
That's a problem D:
The cosine of 8pi is 1! :D that is what my calculator told me lol
Your calculator is quite right...though, if you're allowed to use a calculator, then what's the problem? Why not make it evaluate the cotangent for you? ^.^
Because it gives me a decimal and the answer is looking for a fraction. My calculator won't turn it into a fraction because it has a radical in the fraction.
Let me tell you a little secret... \[\large \cot(x) = \frac1{\tan(x)}\] So if you want the cotangent of something, just divide 1 by its tangent. Go on, give Mr. Calculator another go :)
2.5E12. It won't make fractions :(
ooohh... Okay, what's the tangent, then?
When I put in \[\cos (-8\pi)\div \sin(-8\pi)\]
I get the same answer
tangent of \(-8\pi\) what is it?
2.5E12
tangent. tangent. tangent. :D
Oh mybad! Haha. 4E-13
No... what kind of calculator are you using? D:
TI-84
jeez XD Why don't you just google tan(-8pi) :D
I've tried :( it doesn't have anything for it! That's why I'm here lol. Would you happen to know tan(115pi)
:D
...yes. But I'm not telling you :P You'll need to figure it out for yourself... but I will help you :) Trust me, google tan(-8pi) the answer is simpler than any fraction you could think of...
Oh my gosh...It's 0...so the answer is undefined because it would be 1/0!
Exactly :P See? You CAN do it ^.^
You're smarter than your bloody TI-84 LOL
I have no clue why it's giving my crazy answers.
Don't trust it for now. Want to know another secret that involves your precious tan(115pi) ? ^.^
Yes! I do enjoy a good secret!
Well then, now that we know that Google is a bit more reliable than Mr. TI-84, why don't you ask google the following: tan(0pi) tan(1pi) tan(2pi) tan(3pi) I'll wait ^.^
They're all 0! So tan(xpi) will always be zero! Awesome to know! Why is that? If it is a lot to explain then don't worry about it; I'll just google it.
Correction: tan(xpi) will be zero AS LONG as x is an integer. Don't forget details, silly :P
But yes, you have made a good observation, and the explanation isn't long at all ^.^
And here's why: \[\Large \sin(n\pi)= 0\] for as long as n is an integer and... \[\Large \cos(n\pi) = (-1)^n\] That said, we have \[\Large \tan(n\pi) = \frac{\sin(n\pi)}{\cos(n\pi)}= \frac0{(-1)^n}=0\]
I'm glad you didn't ask Google what tan(115pi) is, because somehow, at that point, Google becomes just as delusioned as TI-84 lawlz ^.^
Yeah! I typed it into google and it gave me something way off. Thanks a ton! You're a life savior!
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