Question

What is t in tan(t)=2 ?

Im using a TI82 but it gives me a weirds numbers.

Answer 1

i know even and odd numbers; whats a wierd number?

Answer 2

t = @12*\[\pi\]-tan^-1(1/2) + pi/2

Answer 3

\[e^{-\pi i \text{weird}}\]

Answer 4

http://en.wikipedia.org/wiki/Weird_number

Answer 5

It doesnt make sense

Answer 6

have you tried replacing the batteries?

Answer 7

are you in radian mode?

Answer 8

Yes Im in radian mode

Answer 9

if the radiation badge has turned colors, its time to get a new ti82

Answer 10

http://www.wolframalpha.com/input/?i=arctan%282%29

Answer 11

My TI89 (It's 89) is new

Answer 12

Can anyone solve t for me?

Answer 13

then you should get something like 1.107 rads

Answer 14

you did solve for t... you have the answer

Answer 15

No I know it's not the answer....

Answer 16

lol...it is

Answer 17

it's just not a form you like

Answer 18

\[tan^{-1}(x) =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\frac{x^{13}}{13}... \]
just plug in x=2

Answer 19

thats the way we had to do it back in the monastary

Answer 20

I get it thanks

Answer 21

satellite73 i need your help on the last problem you helped me with.

Answer 22

unless you specify the domain you are working over you will have an infinite number of solutions...just as your calculator indicated

Answer 23

It's part of this question
Find the maximum and minimum values of
f(x,y) = x+2y on the disk x^2+y^2 ≤1

I have done this for now:

f_1(x,y) = 1
f_2(x,y) = 2

x=cos(t) and y=sin(t)

I have that g(t) = x(t) + 2*y(t) --> g(t) = cost(t) + 2*sin(t)

g'(t) = 0 = 2*cost-sin(t)

Then I can see that:

2cos(t)/cos(t) -sin(t)/cos(t) = 0/cos(t) --> tan(t) = 2

From this point I have no idea what to do.

Answer 24

use lagrange multipliers

Answer 25

That is the given solution....I just cant follow what they are doing

Answer 26

or...just compute \[\cos(\tan^{-1}(2))+2\sin(\tan^{-1}(2))=\sqrt{5}\]

Answer 27

Thank you a lot you just helped me finish this assignment!
Greetings from Denmark

Answer 28

In case anyone is interested, though I'm not sure it sheds much light.