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

- anonymous

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

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- schrodinger

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- jim_thompson5910

Can you post a screenshot of the full problem?

- anonymous

yeah, just one moment

- anonymous

sorry my computer is super slow

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## More answers

- anonymous

##### 1 Attachment

- jim_thompson5910

that's fine

- jim_thompson5910

ah, I see now

- jim_thompson5910

so 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

then 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

so it's just the coordinate points ?

- jim_thompson5910

yeah the y coordinates of each point on the graph of y = tan(x)

- jim_thompson5910

2*tan(x) I mean

- anonymous

ugh thank you, you're a life saver

- jim_thompson5910

you're welcome

- anonymous

wait how'd you get \[\pi/4 \] to be 1

- jim_thompson5910

tan of pi/4 is 1

- jim_thompson5910

if 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

because of that and because tangent = sine/cosine, the two equal values divide to 1

- anonymous

oooooooh okay, i get it now

- jim_thompson5910

I'm glad it's making more sense

- anonymous

Yeah I'm the worst at trigonometry lol

- jim_thompson5910

I'm sure with more practice, you'll get better at it

- anonymous

this is my third year learning this and I still haven't learned it :/

- jim_thompson5910

then a different approach is needed

- anonymous

yes, badly
so for pi/3 , would it be radical 3 ? or do the 2's not cancel out ?

- jim_thompson5910

the 2s will cancel leaving sqrt(3), correct

- anonymous

Okay, I just wanted to make sure cause I've seen it where the 2's aren't being canceled out

- anonymous

and since I'm doubling it, would it be \[2\sqrt{3}\] or ?

- jim_thompson5910

I'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

yes, \[\Large 2\tan\left(\frac{\pi}{3}\right) = 2\sqrt{3}\]

- anonymous

Okay, 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

- jim_thompson5910

hmm, strange

- anonymous

but it just makes sense to cancel them out to me so I figured whoever made that table just forgot that step maybe?

- jim_thompson5910

that's possible

- jim_thompson5910

even teachers who write the problems and examples make typos

- anonymous

I know many lol. But anyways, thank you so much

- jim_thompson5910

sure thing

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