Solving for x in an equation involving trigonometric functions

[dilasluis]

Angle panic!

When we have a relation like, for instance,

f(\theta) + g(\theta) = constant where \theta is an angle, does it hold for any angle such that we can do f(0 \deg) + g (0 \deg) = 0 and we would obtain an universal result? I mean, imagine f(\theta, x) = x \sin \theta, then

x = - \frac{g (0 \deg)}{\sin 0\deg} = - \frac{g (45 \deg)}{\sin 45\deg}

 

[DonAntonio]

When we have a relation like, for instance,

f(\theta) + g(\theta) = constant where \theta is an angle, does it hold for any angle such that we can do f(0 \deg) + g (0 \deg) = 0 and we would obtain an universal result? I mean, imagine f(\theta, x) = x \sin \theta, then

x = - \frac{g (0 \deg)}{\sin 0\deg} = - \frac{g (45 \deg)}{\sin 45\deg}

I've read the above 4 times (the last two rather slowly and carefully) and I still cannot understand what it means...

One thing is sure, though: \sin 0 = 0\\, so it cannot appear in the denominator.

DonAntonio

 

[dilasluis]

Sorry, bad choice on the sine...

But imagine that you have the following equation:

x \cos \theta - y(\theta) \cos^2\theta = 0

I wish to find a solution for x. What I want to know is if it is equivalent:

x = y (0 \deg) = \frac{y (45 \deg) \frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{ y (a \deg) \cos^2 a}{ \cos a}.

Because, in my mind, it should be! If it holds for an unknown \theta, it should hold for any \theta!

 

[dilasluis]

I've read the above 4 times (the last two rather slowly and carefully) and I still cannot understand what it means...

One thing is sure, though: \sin 0 = 0\\, so it cannot appear in the denominator.

DonAntonio

That sine actually made me think on something, even if it holds, the equation would not be valid for that particular value of \theta.

 

[HallsofIvy]

Sorry, bad choice on the sine...

But imagine that you have the following equation:

x \cos \theta - y(\theta) \cos^2\theta = 0

I wish to find a solution for x.

Then you can simply write x= y(\theta)cos^2(\theta)/cos(\theta)= ycos(\theta) as long as cos(\theta) is not 0- that is if \theta is not an odd multiple of \pi/2.

What I want to know is if it is equivalent:

x = y (0 \deg) = \frac{y (45 \deg) \frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{ y (a \deg) \cos^2 a}{ \cos a}.

Because, in my mind, it should be! If it holds for an unknown \theta, it should hold for any \theta!

Again, it is not clear what you mean. What do you mean by "holds for an unknown"?
x^3= 3 for some "unknown" value but does not hold for all x.

 

[dilasluis]

I don't know if you will understand... but I want to know if any value for the angle, as long as it has a finite result, can be used to solve an equation with unknown angles.

How would you solve this, for instance:

x \sin \theta + \sqrt{x \cos \theta} = 0 ?

Could you assume any value for \theta and the resulting x would be the same?

 

[haruspex]

I don't know if you will understand... but I want to know if any value for the angle, as long as it has a finite result, can be used to solve an equation with unknown angles.

How would you solve this, for instance:

x \sin \theta + \sqrt{x \cos \theta} = 0 ?

Could you assume any value for \theta and the resulting x would be the same?

Certainly not. The above equation makes x and θ functions of each other. x = 0 is always a solution, and otherwise it can be simplified to x = cosec(\theta) cot(\theta)

 

[dilasluis]

I'm just writing equations for the sake of it... I'm not trying to solve them...

And would this make sense: If I would make an average of the equation and solve for x, like this:

Suppose you have: x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta = 0

then to find a solution to x could you average the above equation on \theta (NOTE: The above equation is just for illustrative purposes... I don't need help solving it!)

\frac{1}{2 \pi} \int_0^{2 \pi} x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta d\theta = 0.

 

[haruspex]

I'm just writing equations for the sake of it... I'm not trying to solve them...

And would this make sense: If I would make an average of the equation and solve for x, like this:

Suppose you have: x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta = 0

then to find a solution to x could you average the above equation on \theta (NOTE: The above equation is just for illustrative purposes... I don't need help solving it!)

\frac{1}{2 \pi} \int_0^{2 \pi} x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta d\theta = 0.

I'm not at all sure I understand what you're trying to do.
If you have an equation x\cos\theta - x^2 \sin^2\theta - \csc\theta\cot\theta = 0, what do you mean by "finding a solution for x"? Normally the meaning would be to find a function of θ which yields the value of x for any given θ. Or sometimes there might be a specific θ that you are interested in and only want the value of x that goes with that. On yet other occasions (usually re Diophantine equations) you just want any pair x and θ that satisfies the equation.
There may be yet other situations where you want to know the average value of x as θ goes through some range (you would need specify the weight given to each part of that range), but this would not be described as "finding a solution for x".