Question

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

Answer 1

Can you post a screenshot of the full problem?

Answer 2

yeah, just one moment

Answer 3

sorry my computer is super slow

Answer 4

that's fine

Answer 5

ah, I see now

Answer 6

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

Answer 7

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

Answer 8

so it's just the coordinate points ?

Answer 9

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

Answer 10

2*tan(x) I mean

Answer 11

ugh thank you, you're a life saver

Answer 12

you're welcome

Answer 13

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

Answer 14

tan of pi/4 is 1

Answer 15

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)

Answer 16

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

Answer 17

oooooooh okay, i get it now

Answer 18

I'm glad it's making more sense

Answer 19

Yeah I'm the worst at trigonometry lol

Answer 20

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

Answer 21

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

Answer 22

then a different approach is needed

Answer 23

yes, badly

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

I'd have to see the problem (where you didn't see the 2s cancel), but you should have this

\[\Large \tan\left(\theta\right) = \frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}\]
\[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)}\]
\[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}/2}{1/2}\]
\[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\times\frac{2}{1}\]
\[\Large \tan\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{\cancel{2}}\times\frac{\cancel{2}}{1}\]
\[\Large \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\]

Answer 28

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

Answer 29

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

Answer 30

hmm, strange

Answer 31

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

Answer 32

that's possible

Answer 33

even teachers who write the problems and examples make typos

Answer 34

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

Answer 35

sure thing