Finding Tan without a calculator

[lorelyi328]

I'm learning physics, and I'm reviewing over the trig part of the chapter. My professor does not allow us to use calculators. I'm looking at the inverse of the trig functions. What I don't understand is : it says that the reciprocal does not equal the inverse. like tan -1 does not equal 1/tan. So how do you figure this out without a calculator. I was looking also to see if cos would be hyp/adj instead of adj/hyp but this also is not the way it is in the book. Will someone help me ?

Maybe this is a better example. There is an example in my book.

A lakefront drops off gradually an an angle theta. For safety reasons it is necessary to know the depths of the lake at various distances from the shore. To get this information a lifeguard rows out from the shore a distance of 14.0m and drops a weighted fishing line. By measuring the length of the line he determines the depth to be 2.25m what is the value of thea ?

so it shows the answer is theta = tan^(-1) (2.25m/14.0m) = 9.13 degrees.

How would you have known enough to use the inverse ? Why couldn't you just use tan instead of the inverse ? it's hard to explain what the picture looks like, but is it because the surface of the lake is given and considered the adjacent rather than the hypotenuse ?
ThaNKs !

 

[jena]

Hi,

You could try the unit circle method by memorizing it

http://http://www.spsu.edu/math/edwards/1113/inverse.htm"

I think the way that you were trying to find inverse of sine as (1/(sine)) because that gives you cosecant.

I hope at least this bit of info will help

 

[Tide]

Read \sin^{-1}x as "the angle whose sine is x" and similarly for the other inverse functions. That should help.

 

[moose]

Read \sin^{-1}x as "the angle whose sine is x" and similarly for the other inverse functions. That should help.

Yes!
Like if you had \sin\theta = .5

than to get \theta you could plug into your calculator

\sin^{-1}.5

and it would tell you 30 (in degrees)

 

[Ouabache]

you could plug into your calculator
\sin^{-1}.5
and it would tell you 30 (in degrees)

your method is right; only one problem, the professor doesn't allow calculators (see first post)..

I know! Its a trick question... The prof didn't say you could not use a trig table. Take the tan as you did and look up the corresponding angle on the table. You may want to brush up a little on interpolation.

 

[lorelyi328]

okay, so how about, if you have the lengths, but need the angle you use the inverse. if you have the angle then you need to use just cos, sine or tan ? is that right ? I'm not very good with trig. the link did not work, sorry.

 

[jena]

Sorry about the link
http://www.spsu.edu/math/edwards/1113/unitcircle.htm

http://www2.spsu.edu/math/edwards/1113/inverse.htm

Try both, they work now.

 

[lorelyi328]

I love the unit circle links ! you're a GOD !

 

[HallsofIvy]

How would you have known enough to use the inverse ? Why couldn't you just use tan instead of the inverse ?

Tangent takes you from the angle to the ratio. In the problem you cite, you are told the ratio ("To get this information a lifeguard rows out from the shore a distance of 14.0m and drops a weighted fishing line. By measuring the length of the line he determines the depth to be 2.25m") and are asked to find the angle theta.

In other words, instead of going from the angle to the ratio, you want to go from the ratio to the angle. That's the whole point of "inverse" functions: they "go" the opposite way. If y= f(x) then x= f^-1(y).

it says that the reciprocal does not equal the inverse. like tan -1 does not equal 1/tan.

Yes, that's true. It's just an unfortunate notation. Working with numbers a -1 exponent means reciprocal, but with functions, including trig functions, it is used to mean the "inverse" function.

So how do you figure this out without a calculator.

In general, you don't. There are a few values, which others have mentioned here, for which the solution is not too difficult, but for most there is no simple way to find tan^-1(y) (or tan(x) or sin(x), etc. for that matter). Those of us who remember years B.C. (before calculators) looked them up in tables. The tables were themselves created using complicated methods such as Taylor's series or the "CORDIC" algorithm and adding machines!

 

[Ouabache]

In general, you don't. Those of us who remember years B.C. (before calculators) looked them up in tables. The tables were themselves created using complicated methods such as Taylor's series or the "CORDIC" algorithm and adding machines!

Don't schools teach how to use trig tables in case you can't find a calculator? or have access to web (see web calculator) ?

Here is how you can find the angle using the table referenced in my last post:
Your ratio (opposite side length/adjacent side length) = 2.25m/14.0m = 0.160714 (you already figured that out)
From the table this ratio (look under Tan), falls between 9deg and 10 deg.
How do I find degrees to more accuracy?
deg = 9, tan x = 0.1584
deg = y , tan x = 0.160714
deg = 10, tan x = 0.1763

Interpolation primer:
1) on left side, your solution y is at distance y from 9 (lower boundary)(i)
2) on right side take the difference between your ratio and lower boundary (0.160714-0.1584) = 0.002314 (ii)
3) what is total difference between boundary values (left side) 10 - 9 = 1 (iii)
4) what is total difference between boundary values right side = 0.1763 - 0.1584 = 0.0179 (iv)
5) you now have proportion y / 1 = 0.002314 / 0.0179 = 0.12927 ( or i / iii = ii / iv,
where small roman numerals correspond to bolded values found in steps 1 thru 4).
6) add this to the lower boundary value (left side) = 9 + 0.12927 = 9.12927 deg or 9.13 deg
7) to two decimal places, this is the same value you find using a calculator, 9.13 deg