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".