Question

Hi, I'm having trouble simplifying this equation and am stuck after solving the inner tan parts: (2tan(pi/12))/(1-tan(pi/12))

Answer 1

this is how far I got:
\[(2(2-\sqrt(3)))/(\sqrt(3)-1)\]

Answer 2

Better find an identity that is equal to:
\[\frac{2tanx}{1-tanx}\]

Answer 3

Are you sure it is not 1-tan^2(x) on the bottom?

Answer 4

yes I'm sure mertsj, it's 1-tan(pi/12) in the denominator

Answer 5

and I can't think of any identity that would match that, I've looked at my identity's page

Answer 6

I typed it up a little more prettier: \[2\tan(\pi/12)/1-\tan(\pi/12)\]

Answer 7

Wolfram Alpha says it should reduce to \[\sqrt{3} -1\]

Answer 8

http://www.wolframalpha.com/input/?i=%282tan%28pi%2F12%29%29%2F%281-tan%28pi%2F12%29%29

Answer 9

Did you try going back to sin and cos?

Answer 10

No, not yet mertsj, I was thinking that'd be too cumbersome

Answer 11

I'd be especially confused on what to do in the denominator then, 1 - sin/cos(pi/12) ? Do both the sin and cos take pi/12 then or what?

Answer 12

2sin(pi/12)]cos(pi/12)-sin(pi/12)

Answer 13

bunch of half angle formula right?

Answer 14

mertsj: is that bracket after the first sin(pi/12) supposed to be a division sign?

Answer 15

yes

Answer 16

satellite: I'm not exactly sure, just some simplification problems with trig functions

Answer 17

what is wrong with your first answer? looks ok to me

Answer 18

I don't think it's simple enough, I am wondering how to go from that to \[\sqrt{3} -1 \]

Answer 19

\[\frac{2(2-\sqrt{3})}{\sqrt{3}-1}\] maybe rationalize the denominator by multiplying by
\[\sqrt{3}+1\] top and bottom

Answer 20

your denominator will be 2, and that will cancel with the 2 in the numerator
so last job is
\[(2-\sqrt{3})(\sqrt{3}+1)\]

Answer 21

yes that is where I am. Do you mean to multiply by \[\sqrt{3} - 1 instead of \sqrt{3} + 1 \] ?

Answer 22

no

Answer 23

but how will the one with the plus cancel out the \sqrt(3) - 1 in the denominator ?

Answer 24

if you want to rationalize the denominator, multiply by \(\sqrt{3}+1\) because then yo will get
\[(\sqrt{3}-1)(\sqrt{3}+1)=3-1=2\]

Answer 25

as \((a-b)(a+b)=a^2-b^2\)

Answer 26

oh ok

Answer 27

and as i said the 2 in the denominator will cancel with the 2 in the numerator, then multiply
\[(2-\sqrt{3})(\sqrt{3}+1)\] and i bet you get your answer

Answer 28

Everytime I get \[\sqrt{3} + 1\] instead of the \[\sqrt{3}-1\] answer that I got from wolfram alpha

Answer 29

really?

Answer 30

yes, I tried it multiiple times

Answer 31

minus square root of 3 times square root of 3 is -3 and 2 times 1 is 2 and 2 - 3 = -1

Answer 32

\[(2-\sqrt{3})(\sqrt{3}+1)\]
\[2\sqrt{3}+2-\sqrt{3}\sqrt{3}-\sqrt{3}\]
\[2\sqrt{3}+2-3-\sqrt{3}\]
\[\sqrt{3}-1\]

Answer 33

Ah, I didn't completely finish distributing the first F in FOIL that is why I missed the 2 from 2 minus sqrt 3 times sqrt 3. Thank you so much