# Trig Equation — help to solve please: 0=cos^2(t)−t sin^2(t)

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Homework Statement:
##0 = \cos^2(t) - t\sin^2(t)##
Relevant Equations:
N/A
I'd like to solve ##0 = \cos^2(t) - t\sin^2(t)## but it's been forever since I've done some trig and I'm real rusty.

I've tried rewriting terms using identities such as ##\sin^2(t) = 1 - \cos^2(t)## but not getting anything helpful. Can I get a point in the right direction?

## Answers and Replies

PeroK
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Homework Statement:: ##0 = \cos^2(t) - t\sin^2(t)##
Relevant Equations:: N/A

I'd like to solve ##0 = \cos^2(t) - t\sin^2(t)## but it's been forever since I've done some trig and I'm real rusty.

I've tried rewriting terms using identities such as ##\sin^2(t) = 1 - \cos^2(t)## but not getting anything helpful. Can I get a point in the right direction?

The presence of ##t## makes that a transcendental equation. It's only solvable numerically.

• opus
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You're talkinga bout the variable t preceding the second term, correct?

PeroK
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You're talkinga bout the variable t preceding the second term, correct?
I'm talking about a mixture of trig and linear functions of ##t##.

• opus
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Ok wow I guess I should rephrase my question then, since I thought this was going to be a simplish trig equation. This trig equation is a part of a bigger problem where I am trying to find out what values of t make a position vector and an acceleration vector perpendicular. I've attached a pic of the LaTeX'd problem so far.
I guess my question now is, since it can only be solved numerically, what would this mean for this problem?

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PeroK
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Ok wow I guess I should rephrase my question then, since I thought this was going to be a simplish trig equation. This trig equation is a part of a bigger problem where I am trying to find out what values of t make a position vector and an acceleration vector perpendicular. I've attached a pic of the LaTeX'd problem so far.
I guess my question now is, since it can only be solved numerically, what would this mean for this problem?

Try expressing the equation in terms of ##\tan t## and you can identify where the solutions are from a sketch.

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Wouldn't I run into problems if I did the following to get tan(t)? :

$$t\sin^2(t) = \cos^2(t)$$
$$t\tan^2(t) = 1$$
$$\tan(t) = \sqrt{\frac{1}{t}}$$

The problems I see with this are dividing by ##\cos^2(t)## as it could be zero. Also, t in the denominator on the final line could be zero as well.

PeroK
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The technique is to look separately for solutions where ##\cos t =0##.

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Ok so spent some time trying to graph this thing and wasnt having much luck. Went to Desmos to have a look and it's an absolute nightmare of a graph. Also tried to express ##\tan(t) = \sqrt{\frac{1}{t}}## a different way but of course that gave me the same thing in Desmos. This is assuming ##\cos(t) \neq 0##.
Should I keep going down this route with my equation as is, or find something else?

Edit: May have something, standby.

PeroK
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You should be able to sketch ##\tan t## and ##\sqrt{1/t}## without software.

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Yes indeed, I was trying to graph one function instead of two separate ones not sure why. Ive got the sketch now though.
Alright I think there are infinite solutions to ##\tan(t) = \sqrt{\frac{1}{t}}## and I see this with the sketch.
Now, these values are not pretty so I'm not sure how to represent what I have here with the original problem.

That is, we had ##0 = t(\cos^2(t)-t\sin^2(t))## and t=0 is one solution. We know there are more solutions by what we just did with the sketch, but how do we use what we just did?

PeroK
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Yes indeed, I was trying to graph one function instead of two separate ones not sure why. Ive got the sketch now though.
Alright I think there are infinite solutions to ##\tan(t) = \sqrt{\frac{1}{t}}## and I see this with the sketch.
Now, these values are not pretty so I'm not sure how to represent what I have here with the original problem.

That is, we had ##0 = t(\cos^2(t)-t\sin^2(t))## and t=0 is one solution. We know there are more solutions by what we just did with the sketch, but how do we use what we just did?

You can see that the solutions are close to ##n \pi##, especially for large ##t##.

Last edited:
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Im not seeing that at all. For example at n = 1, we should have ##\frac{3\pi}{2}## its a little over ##\frac{\pi}{4}## as I'm looking at Desmos for better accuracy.

PeroK
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Im not seeing that at all. For example at n = 1, we should have ##\frac{3\pi}{2}## its a little over ##\frac{\pi}{4}## as I'm looking at Desmos for better accuracy.

You're right, I've corrected it.

What I meant was there is one solution for each zero of the ##\tan## function, which occur at ##t = n\pi##, ##n = 0, 1, 2 \dots##.

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Ok now I definitely can see that. So then would it be satisfactory to say that, with respect to the original problem, ##\cos(\theta) = 0## when t = 0 and when ##t = (n + \frac{1}{2})\pi## where n is an integer?

PeroK
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Ok now I definitely can see that. So then would it be satisfactory to say that, with respect to the original problem, ##\cos(\theta) = 0## when t = 0 and when ##t = (n + \frac{1}{2})\pi## where n is an integer?

There is one solution for every zero of the ##\tan## function. Plus the solution ##t = 0## itself. The solutions tend to ##n\pi## as ##n## increases, but they are never equal to ##n \pi##.

Note: I've corrected the formual to ##n \pi##.

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If we look at the trigonometric equation in a vacuum, I can see how we can say that the solutions tend to ##n\pi##. I also can see that there are an infinite amount of solutions.
But the original question is to find when the vectors are perpendicular (that is, when is cos = 0). So it seems I have an infinite amount of solutions and no way to express them. Because as you say, then TEND to ##n\pi## but they must be exact for the vectors to be perpendicular.

Gold Member For reference, these points of intersection are all solutions to the equation (in other words, when my two vectors are perpendicular). But am I correct if I say that there is no explicit way to state every single solution here?

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PeroK
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That's right, you just label your solutions ##t_n## or whatever and refer to them that way. ##t_n \rightarrow n \pi## for large ##n## is quite an important point. It means the zeroes quickly become almost periodic.

• opus
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Ok thank you kindly!

Merlin3189
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So I wonder what is the physical situation giving rise to the equation?

It feels like an Archimedian spiral and a vector from a non-central point, but I'm not a mather, so that's just a wild guess.

neilparker62
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