Proving There's No Solution to csc(x)+cot(x)=1

[e^(i Pi)+1=0]

solve this equation
csc(x)+cot(x)=1

after some symbol shunting this simplifies to sin-cos=1

I believe there is no solution to this problem. Let sin = a and cos = b

if a-b=1

then

a²+b²=1+2ab

but we know that a²+b² must =1 therefor, no solution. Is this a valid proof?

 

[Dick]

solve this equation
csc(x)+cot(x)=1

after some symbol shunting this simplifies to sin-cos=1

I believe there is no solution to this problem. Let sin = a and cos = b

if a-b=1

then

a²+b²=1+2ab

but we know that a²+b² must =1 therefor, no solution. Is this a valid proof?

Not at all, a=1 and b=0 solves a-b=1.

 

[e^(i Pi)+1=0]

Well then I am stumped. Are you saying there is a solution? It certainly doesn't look it when I graph it.

 

[QuarkCharmer]

Of course there is a solution. What's the max value that sin(x) or cos(x) will output? Is there any angle x, such that sin(x) or cos(x) will output 1, and the equation sin(x) - cos(x) = 1 is still satisfied?

Clearly from that equation, you know that either:
a.) cos(x) outputs a value >0, and then to satisfy, sin(x) must output a value >1.
or...

 

[Mentallic]

Well then I am stumped. Are you saying there is a solution? It certainly doesn't look it when I graph it.

You must be graphing it incorrectly then. Think about values of x where sin(x) is positive and cos(x) is negative. Can't it certainly be possible that sin(x)-cos(x)>1 in this case?

To begin solving this problem, try converting sin(x)-cos(x) into $R\sin(x+\alpha)$ for some constants R and $\alpha$ that you need to determine.

 

[e^(i Pi)+1=0]

Of course there is a solution. What's the max value that sin(x) or cos(x) will output? Is there any angle x, such that sin(x) or cos(x) will output 1, and the equation sin(x) - cos(x) = 1 is still satisfied?

Well when you put it like that it's pretty damn obvious :shy:

Graphing $\frac{1}{sin(x)}$+$\frac{1}{tan(x)}$ yields holes at y=1 and y=-1. What other way is there to graph it?

 

[HallsofIvy]

What does that graph have to do with anything? In particular, what does tan(x) have to do with anything? You are asking about the equation csc(x)+ cot(x)= 1. At $x= k\pi/2$, csc(x)= 1 and cot(x)= 0. THAT'S what you should be graphing.

 

[Villyer]

What does that graph have to do with anything? In particular, what does tan(x) have to do with anything? You are asking about the equation csc(x)+ cot(x)= 1. At $x= k\pi/2$, csc(x)= 1 and cot(x)= 0. THAT'S what you should be graphing.

He was graphing the original equation in order to see where it equals 1.

 

[e^(i Pi)+1=0]

What does that graph have to do with anything? In particular, what does tan(x) have to do with anything? You are asking about the equation csc(x)+ cot(x)= 1. At $x= k\pi/2$, csc(x)= 1 and cot(x)= 0. THAT'S what you should be graphing.

$\frac{1}{tan}$ is cotan, and until my TI-84 magically grows cot, sec and csc buttons that's how I'll have to graph it.

 

[Dickfore]

If you know half angle formulas, then your original equation reduces to:
$$\csc x + \cot x = \frac{1 + \cos x}{\sin x} = \frac{2 \, \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} = \cot \frac{x}{2}, \ \cos \frac{x}{2} \neq 0$$
You may check that when:
$$\cos {\frac{x}{2} } = 0 \Leftrightarrow \frac{x}{2} = \frac{\pi}{2} + n \, \pi \Rightarrow x = (2 n + 1) \pi$$
then
$$\sin x = \sin \left[ (2 n + 1) \pi \right]$$
so there is no loss of solutions by canceling with $\cos ( x/2 )$. On, the other hand, if you multiply the equation by $\sin x$, and convert it as you did to:
$$\sin {x} - \cos{x} = 1$$
you may see that $x = (2 n + 1) \pi$ are solutions of this equation, whereas they are not for the original equation.

 

[e^(i Pi)+1=0]

I can tentatively follow your logic, but I'm still not clear on what the solution is. Is the answer x= $\frac{∏}{2}$ [0,2∏) or no solution? If my simplification generated an invalid answer then surely yours did as well because neither works in the original equation.

 

[Dickfore]

$$\cot \frac{x}{2} = 1$$
has the solution:
$$\frac{x}{2} = \frac{\pi}{4} + n \pi \Rightarrow x = \frac{\pi}{2} (4 n + 1)$$The set of solutions:
$$(2 n + 1) \pi = (4 n + 2) \frac{\pi}{2}$$
is not of the above form, so do not need to exclude any of them.

 

[e^(i Pi)+1=0]

That doesn't answer my question.

 

[Bohrok]

What was your question?
Also, I'm not sure about x=∏/2 [0,2∏) a couple posts ago.

Dickfore did get the correct general solution of $x=\frac{\pi}{2}(4n+1)$

I also managed to rewrite the original equation as sinx - cosx = 1, but this has one set of solution, x = 2πn + π, which doesn't work for the original equation.

 

[e^(i Pi)+1=0]

Again, my question is..

If my simplification generated an invalid answer then surely yours did as well because neither works in the original equation.