Dividing by trigonometric functions

In summary, the individual is trying to solve a physics problem involving projectile motion. They have set up their equations and are stuck on how to express time (t) using other known equations. They are aware that dividing by a trigonometric function, specifically cosine, can be invalid since it can equal zero. However, they cannot see any other way to express time. They are seeking clarification on whether their approach is valid and if there are any exceptions to the rule about dividing by trigonometric functions. They have also mentioned finding a similar question on Yahoo Answers and are seeking the opinion of the Physics Forums community.
  • #1
frozonecom
63
0
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.
 
Mathematics news on Phys.org
  • #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:
 
  • #3
frozonecom said:
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.

[/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
 
  • #4
frozonecom said:
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.
[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?)

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.
 
  • #5


Hello,

Thank you for reaching out to us with your question. It is understandable that you are confused about dividing by trigonometric functions in this case.

In general, dividing by a trigonometric function is not a valid algebraic move. As you mentioned, this is because the trigonometric function can be equal to zero, which would make the division undefined.

However, there are some exceptions to this rule. For example, if the trigonometric function is squared, then it can be divided by. In your case, if you have a trigonometric function squared, you can take the square root of both sides of the equation to eliminate the squared function and then divide.

Another exception is if the trigonometric function is part of a larger expression, and dividing by it would not cause the overall expression to be undefined. In your case, you have a constant (270) on one side and a trigonometric function (cosθ) multiplied by a variable (t) on the other side. Dividing by cosθ would not make the overall expression undefined, so it would be a valid algebraic move.

In summary, dividing by a trigonometric function is generally not a valid algebraic move, but there are exceptions to this rule. In your case, it would be valid to divide by cosθ since it is part of a larger expression and would not make the overall expression undefined. I hope this helps clarify your confusion. If you have any further questions, please let me know.
 

FAQ: Dividing by trigonometric functions

1. What are the basic rules for dividing by trigonometric functions?

The basic rule for dividing by trigonometric functions is to multiply the numerator and denominator by the reciprocal of the trigonometric function in the denominator. For example, if we have the expression sin(x)/cos(x), we can multiply both the numerator and denominator by 1/sin(x) to get 1/cos(x).

2. Can we divide by 0 when dealing with trigonometric functions?

No, we cannot divide by 0 when dealing with trigonometric functions. This is because the values of trigonometric functions, such as sine and cosine, become undefined at certain points, including when the input angle is 90 degrees. Dividing by 0 would result in an undefined answer.

3. How do we simplify expressions involving division of trigonometric functions?

To simplify expressions involving division of trigonometric functions, we can use trigonometric identities, such as the Pythagorean identities, to rewrite the expression in a simpler form. We can also use the basic rules for dividing by trigonometric functions and simplify further from there.

4. Can we divide by trigonometric functions in radians and degrees?

Yes, we can divide by trigonometric functions in both radians and degrees. However, we must be careful when converting between the two units as they have different values for trigonometric functions. It is important to use the correct unit of measurement when solving trigonometric equations involving division.

5. Are there any special cases when dividing by trigonometric functions?

Yes, there are some special cases when dividing by trigonometric functions. One example is when dealing with tangent, which becomes undefined when the input angle is 90 degrees. Another special case is when dividing by sine or cosine, which can result in an undefined answer when the input angle is 0 degrees or 180 degrees. It is important to be aware of these special cases and handle them carefully when dividing by trigonometric functions.

Similar threads

Replies
9
Views
1K
Replies
14
Views
3K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
2
Views
2K
Replies
7
Views
3K
Back
Top