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anonymous
 one year ago
y = 2tan(theta) . Find the value of tangent, then double it.
It's a table with pi/6 through 2pi
anonymous
 one year ago
y = 2tan(theta) . Find the value of tangent, then double it. It's a table with pi/6 through 2pi

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Can you post a screenshot of the full problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, just one moment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry my computer is super slow

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1ah, I see now

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so what you do is replace \(\Large \theta\) (greek letter theta) with the numbers in the top row Use the unit circle or a table to find that \[\Large \tan(\theta) = \tan(0) = 0\] \[\Large \tan(\theta) = \tan\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{3}\] \[\Large \tan(\theta) = \tan\left(\frac{\pi}{4}\right) = 1\] etc etc

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then you double each result \[\Large 2*\tan(\theta) = \tan(0) = 2*0 = 0\] \[\Large 2*\tan(\theta) = 2*\tan\left(\frac{\pi}{6}\right) = 2*\frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}\] \[\Large 2*\tan(\theta) = 2*\tan\left(\frac{\pi}{4}\right) = 2*1 = 2\] etc etc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it's just the coordinate points ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah the y coordinates of each point on the graph of y = tan(x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.12*tan(x) I mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ugh thank you, you're a life saver

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait how'd you get \[\pi/4 \] to be 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1tan of pi/4 is 1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1if you look at the unit circle, you'll find that sine and cosine have the same value at pi/4 sin(pi/4) = cos(pi/4)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1because of that and because tangent = sine/cosine, the two equal values divide to 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oooooooh okay, i get it now

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm glad it's making more sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I'm the worst at trigonometry lol

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'm sure with more practice, you'll get better at it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is my third year learning this and I still haven't learned it :/

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1then a different approach is needed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, badly so for pi/3 , would it be radical 3 ? or do the 2's not cancel out ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the 2s will cancel leaving sqrt(3), correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I just wanted to make sure cause I've seen it where the 2's aren't being canceled out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and since I'm doubling it, would it be \[2\sqrt{3}\] or ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1I'd have to see the problem (where you didn't see the 2s cancel), but you should have this \[\Large \tan\left(\theta\right) = \frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}\] \[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)}\] \[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}/2}{1/2}\] \[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\times\frac{2}{1}\] \[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{\cancel{2}}\times\frac{\cancel{2}}{1}\] \[\Large \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes, \[\Large 2\tan\left(\frac{\pi}{3}\right) = 2\sqrt{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, I think I'm getting a little more. I saw it on some website and it was a table like I'm filling out and the 2's weren't canceled out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but it just makes sense to cancel them out to me so I figured whoever made that table just forgot that step maybe?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1that's possible

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1even teachers who write the problems and examples make typos

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know many lol. But anyways, thank you so much
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