Graphing a Hyperbolic Trigonometric Function: Radian vs. Degree Mode

  • Thread starter Thread starter sophzilla
  • Start date Start date
  • Tags Tags
    Calculator
Click For Summary
When graphing the hyperbolic trigonometric function sin(x)cosh(x) - sinh(x)cos(x), using radian mode on the calculator is recommended for accurate results. The default window settings on a TI-83/84 calculator may not display key points like cos(x) hitting zero in degree mode due to range limitations. It's crucial to ensure that both functions are evaluated in the same mode, either degrees or radians, to avoid discrepancies in the graph. The discussion emphasizes that most advanced math courses utilize radians, making it a safer choice for graphing. Consistency in the argument type is essential for accurate graphing outcomes.
sophzilla
Messages
20
Reaction score
0
Hello - I'm sorry if this question doesn't really apply here, but the Math & Science Tutorial wasn't very active so I decided to come here instead...

What should my calculator mode be in when graphing a hyperbolic WITH trigonometric function? (My function is sinx*coshx - sinhx*cosx). I'm getting 2 different graphs between radian and the degree mode.

I think it should be in the radian mode...?

Thanks for your help.
 
Physics news on Phys.org
What kind of calculator are you using?

I'll assume for a minute that you are using a TI-83/84. When you press the "WINDOW" key (next to "Y=") what is your X min, X max, Y min, Y max values?

The default is:
Xmin = -10
Xmax = 10
Ymin = -10
Ymax = 10

Lets say you are plotting cos(x)
Well from basic trig you know that:
cos(x=\frac{\pi}{2}) = 0
and that \frac{\pi}{2}=90
where 90 is in degrees.

So if you have your default Xmin, Xmax, ... values above, then this means that you are allowing the calculator to start at the value of -10 and increment up to 10.

If your calculator is in degree mode you will never see cox(x) hit 0. Do you see why this is? Our range is too small. But if you put your calculator in radian mode, then guess what... \frac{\pi}{2}=1.571 so we are within our range of values between -10 and 10, thus we will see something useful.

Also, if we let y=cos(x). Then, y will range from -1 to 1. So you only need to set
Ymin = -1
and
Ymax = 1
 
Pretty much any math class I have ever taken past high school used radians. I would suggest the same. It really doesn't matter which one you use as long as both functions get the same type of argument, either degrees or radians. Don't mix them.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
5K
Replies
3
Views
2K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 2 ·
Replies
2
Views
9K
Replies
2
Views
11K
Replies
4
Views
2K
Replies
4
Views
9K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K