1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dividing by trigonometric functions

  1. Jul 14, 2013 #1
    Hello. I was doing a (simple) physics problem and stumbled with a mathematical problem.

    I was doing a projectile motions problem and I have set up my equation like this:


    Δx = Vi (cosθ) (t)
    270= 25cosθ t
    t = 270 / (250cosθ)

    And this is where I'm having problems.
    I know from my high school trig that doing division by a trig function is invalid since cosθ can be zero. However, I see no other way to express t (time) with other equations I know.

    So, my main question is, is what I did a valid move algebraically?
    Are there any exceptions to that rule about dividing by trigonometric functions?

    Note: I actually saw a yahooanswer post about this but I really want to know physicsforums's say about this. I know this might come as an easy question for some but it really came very confusing for me.
    Help would be very appreciated.
     
  2. jcsd
  3. Jul 15, 2013 #2
    In the equation 270= 25cosθ t if assume cosθ to be 0, you will get 270=0! This is not possible. ∴cosθ≠0 and hence you can divide cosθ and get an equation for t.

    I think you meant
    t = 270 / (25cosθ) and not t = 270 / (250cosθ):confused:
     
  4. Jul 15, 2013 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    [/QUOTE]

    You do not divide with a cosine function, but with a number. Theta is a well defined angle in case of projectile motion, the angle the projectile was thrown at with respect to the horizontal. It must be 90 (-90) degrees to get cosθ=0 and that corresponds throwing it vertically up or down. There is no horizontal displacement in these cases.

    ehild
     
  5. Jul 20, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [itex]cos(\theta)[/itex] can be 0 but you should be able to realize that in a problem like this, it won't be! [itex]cos(theta)= 0[/itex] for [itex]\theta[/itex] 90 degrees or 270 degrees. I presume your [itex]\theta[/itex] is the angle the trajectory makes with the horizontal so it can't be 270 degrees and 90 degrees means the projective is 'fired' straight up. And, of course, if the projective goes straight up, it will come straight back down- so you could do that case separately. There is no objection to saying "Clearly, if [itex]\theta[/itex] cannot be 90 because then the ball will not have ANY horizontal motion and x can never be 270. If [itex]\theta[/itex] is not 90 degrees, [itex]cos(\theta)[/itex] is not 0 so we can divide by [itex]cos(\theta)".

    (I do have an objection to the "25" in one equation becoming "250" in the next! May we assume that extra "0" is a typo?)

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dividing by trigonometric functions
  1. Dividing Functions (Replies: 3)

Loading...