Question

how do i solve the inverse tangent of 6?

Answer 1

I think this problem actually requires that you have the rest of the problem written. I had an idea on how to solveit, but it requires the entire problem you had xD

Answer 2

lol it looks less scary this way. But Here's the entire equation folks: \[Cos (\frac{ 1 }{ 2 } \tan^{-1} 6)\]

Answer 3

Well, because when you have a problem like this, its actual triangles and not just calculator work. This is a triangle problem, so thats why you dont need a calculator forit. If it were just tan^(-1)6 then its all calculator.

Answer 4

Okay, so we need to say that the whole (1/2)tan^(-1)6 is going to be θ. So we will, eventually, have cos(θ) = ?. So now how to deal with the 1/2arctan bleck. So we need a triangle for this. First, though, if arctan (sorry, I like writing arc better), so if arctan6 = θ, then tanθ = 6. That being said, in terms of a triangle, tangent is opposite over adjacent. So inside of the parenthesis, I have (1/2)tanθ = 6. Solving for tanθ, I have tanθ = 12. Now HERE is where the triangle comes in. *draws*

Answer 5

What we basically have done is said that the cosine of a triangle whose tangent is 6, has an angle of 1/√(145), or basically √(145)/145.

Answer 6

Could be wrong, but I have an idea what to do with it x_x

Answer 7

Wait, wait, wait, wait, I got it!!!

Answer 8

@Psymon thank you so much for that explanation. This actually makes sense- my only concern is that i know i need to use a sum or difference formula for this kind of problem because there is a MUCH simpler problem in the book. This is that problem on steroids *_*

Answer 9

I need to correct myself, I know how to solve this. Can I re-explain? x_x

Answer 10

go ahead! :)

Answer 11

Okay, we need to do a half-angle formula! We need tosay that θ is arctan(6), but we needour cos(θ/2) formula :P

Answer 12

HOW DID I NOT SEE THAT. Man!!

Answer 13

I dont know, not sure how I didnt, haha.

Answer 14

ok ill plug it into the cos\[\cos \theta \] formula and check my answer with you if you don't mind :) I make silly calculator mistakes

Answer 15

Haha, okay. But yeah, I just checked with my calculator. I plugged in the triangle method along with the flat out put it inthe calculator method and I got them both to match :P

Answer 16

did you use the formula \[\sqrt{\frac{ 1+\cos \alpha }{ 2 }}\] ?

Answer 17

i used this one because 6 is in quadrant 4 :p

Answer 18

But cos is positive in quadrant 4 :P

Answer 19

why wouldn't it be positive? O_0

Answer 20

Sorry, I read it wrong :P But yeah, youre right, haha. But its √[(1+cos2θ)/2]

Answer 21

Wait....

Answer 22

Oops, youre rightxD

Answer 23

Got it confused with power-reducing

Answer 24

haha that's ok, so my theta in the half angle formula is arctan 6??

Answer 25

Yes, but from there we need to get the cos of that, which is done with the triangle :P