Question

approximation of tan 28 degrees -_-''

Answer 1

Well it's close to 30 degrees.

Answer 2

Well that's VERY close to tan (30 degrees) which is 1/3

Answer 3

There is less than a 8% diffrerence.

Answer 4

Yes, I know that, i have to use newtons method though

Answer 5

Ohh dear!!!

Answer 6

>.<

Answer 7

Yepp ! its such a pain!

Answer 8

lol

Answer 9

newtons method is a good way to approximate though =]

Answer 10

you mean the \[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]thing?

Answer 11

I can't wait to learn toylor polynomials to approxmitate though.

Answer 12

Yeah that's the one @TuringTest .

Answer 13

Yes @TuringTest

Answer 14

Yeah, the thing that you can simply program into your calculator to have it do it for you automatically

Answer 15

But what if you can't use a calculator? :O .

Answer 16

fill in the blanks with \(x_0=30^{\circ}=\frac\pi6\)

Answer 17

Then you have to do it manually, which can really be a pain.

Answer 18

@Hero I have to do it by hand :/ no calculators! and ok

Answer 19

first off, what's\[f(x_0)\]in our case?

Answer 20

tan x

Answer 21

that is our \(f(x)\) but we are looking at \(f(x)\) near \(x_0=\frac\pi6\)
I asked you what \(f(x_0)=f(\frac\pi6)\) is because that is the first thing we see in our formula

Answer 22

OH thats what the question was. Yes its is pi/6 b/c 30degrees

Answer 23

yeah, so what is tan(pi/6) is what my question amounts to

Answer 24

sqrt3/3

Answer 25

right, so that's one piece of the puzzle
now what about f' ?

Answer 26

1/cos^2(pi/6)

Answer 27

right, that's f'(pi/6), which is what?

Answer 28

1/cos(pi/6)= sqrt3/3 right? but what is it ^2??

Answer 29

cos(pi/6)=(sqrt3)/2

Answer 30

so 1/cos(pi/6)=2/(sqrt3)
what is 2 squared?
what is sqrt3 squared?

Answer 31

4/3?

Answer 32

yes :)
so now you can fill in the blanks to get a linear approximation for f(x) near pi/6

Answer 33

\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)\]you know each part now

Answer 34

so i then do rt3/3+ 4/3(pi/90)= my answer?

Answer 35

you forgot about \(x_0\)

Answer 36

oh I see what you did, but it should be negative because \(x<x_0\)

Answer 37

i thought (xo) was sqrt3/3 ?

Answer 38

yes that is f(x_o), but I am talking about x_o=pi/6

Answer 39

better to call it 30 deg right now

Answer 40

And now you have to use newtons method to approximate the square root of 3?

Answer 41

so pi/6 + sqrt/3 - 4/3(pi/90) ?

Answer 42

so\[f(x)\approx f(x_0)+f'(x_0)(x-x_0)=\frac{\sqrt3}3+\frac43(28^{\circ}-30^{\circ})\]

Answer 43

yes

Answer 44

wait where did that other pi/6 come from?

Answer 45

you said i was missing the xo pi/6 lol confused me... the degrees..do i have to convert them to radians?

Answer 46

yes, because radians have no units, you need a unitless quantity for the value of tangent

Answer 47

so essentially the 28-30 degrees actually is pi/6-pi/90 ?

Answer 48

pi/6=30
28-30=-2 (degrees) =-pi/90 radians

Answer 49

sorry, im confused. how do i know that -2 degrees is pi/90 ?

Answer 50

@jennilalala He converted degrees to radians.

Answer 51

I think...

Answer 52

yep
-2 degrees is -1/180th of a circle, which is 2pi radians
-2pi/180=-pi/90

Answer 53

in general\[x~\text{radians}\times\frac{180}\pi=y~\text{degrees}\]and vice versa

Answer 54

\[y~\text{degrees}\times\frac\pi{180}=x~\text{radians}\]

Answer 55

rt3/3+4/3(-pi/90)= rt3/3-4pi/270 ...??

Answer 56

yes, that looks right :)

Answer 57

so I can leave that as the answer? how do you subtract that ive never even done that before

Answer 58

there isn't much to do, you can't simplify pi or sqrt3. You could get a common denominator if you wanted, but I see little point in that.

Answer 59

i see. thank you very much for this detailed explanation/walkthrough

Answer 60

welcome
you should also always check your answer against the value your calculator gives you for tan(28)
I get 0.5308... for the approximation and 0.5317... for the actual value, so it's pretty darn close :)

Answer 61

I wish that he could just write tan 30 :/ . It's pretty close.

Answer 62

@Dido525 hahahaha me too man. me too...

Answer 63

&you mean *we? as in students...@Dido525

Answer 64

but it's not nearly as close as our approximation, I think if you used tan(30) in place of tan(28) our satellites would be flying off into space :P

Answer 65

I said he :P .

Answer 66

No kiding!!! @TuringTest

Answer 67

that's happened before actually, something like that where scientist used the wrong approximation and something space completely messed up

Answer 68

I remember something like that, a guy in ASA not converting feet to meters. I think I was in high school.

Answer 69

NASA*

Answer 70

The didn't convert units. So they ended up making a space probe which crashed on mars because they didn't convert units properly.