Trigonometric Ratios: Explaining Angle $\theta$

[paulmdrdo1]

can you explain to me what my trigonometry book say

"the value of each ratio is determined only by the angle $\theta$ and not by the particular point $P(x,y)$ on the terminal side; that is , the value of the ratio is a function of the angle $\theta$."

what does "determined only by angle $\theta$ mean? and the phrase "the value of the ratio is a function of the angle $\theta$."

please bear with me. I'm not a native English speaker.

 

[MarkFL]

I believe what they are saying is that when you have two right triangles which are similar, which means the respective ratios of the sides are equal, you will find the trigonometric values of the angles remain the same as well.

Suppose you have a right triangle with an angle $\theta$ subtended by one of the legs (the adjacent leg) and the hypotenuse. Suppose the leg is 1 unit in length and the hypotenuse is 2 units in length. We may then state:

$$\cos(\theta)=\frac{1}{2}$$

Now, consider if we scale this triangle up or down by multiplying all three sides by some factor $0<k$. We will find:

$$\cos(\theta)=\frac{k}{2k}=\frac{1}{2}$$

As you can see the size of the triangle, which correlates to the point on the terminal side, does not matter. It is the ratio of the two sides to one another, not their sizes, that determines the values of the trigonometric functions of that angle.

 

[Evgeny.Makarov]

what does "determined only by angle $\theta$ mean? and the phrase "the value of the ratio is a function of the angle $\theta$."

In general, suppose we have several parameters, such as angle $\theta$, coordinates $x$ and $y$ and ratio $r$. We assume that there is some relationship between these parameters. Thus, it is not true that each of these parameters cannot vary independently of the others. Let us call the sequence of values $(\theta,x,y,r)$ feasible if there exists an actual diagram where the parameters in question take these values. For example, if $\theta$ is the angle at $(0,0)$ in the right triangle with vertices $(0,0)$, $(x,0)$ and $(x,y)$ and $r=y/x$, then the sequence $(\pi/6,\sqrt{3}/2,1/2,1/\sqrt{3})$ is feasible, but $(\pi/6,1,0,0)$ is not. Then we say that $r$ is determined only by $\theta$, or that $r$ is a function of $\theta$, if for all values $x$, $y$, $x'$, $y'$, $r$, $r'$, if both $(\theta,x,y,r)$ and $(\theta,x',y',r')$ are feasible, then $r=r'$. That is, there cannot be two diagrams with the same angle $\theta$, but two different ratios $r$ and $r'$. The same value of the argument cannot correspond to two different values of the function--this is the key property of functions.

 

[paulmdrdo1]

what do you mean by "if all values x, y, x′, y′, r, r′, if both (θ,x,y,r) and (θ,x′,y′,r′) are feasible, then r=r′.

feasible in what sense?

can you explain it in a more comprehensible manner? thanks!

 

[paulmdrdo]

just think about it this way, a unique value of trigonometric ratio is determined whenever $"\theta"$ is given. since we're talking about function here we can treat $"\theta"$ as the "domain"(set of angles) of the function and the value of the ratio as the "range". as evgenymakarov pointed out there can be no same value of the "domain"(angles) which corresponds to two different values of the ratio(range).

 

[Evgeny.Makarov]

what do you mean by "if all values x, y, x′, y′, r, r′, if both (θ,x,y,r) and (θ,x′,y′,r′) are feasible, then r=r′.

Perhaps two ifs in a row was unclear. I was saying the following. Suppose that for all values $x$, $y$, $x'$, $y'$, $r$, $r'$, if both $(\theta,x,y,r)$ and $(\theta,x',y',r')$ are feasible, then $r=r'$. Then we say that $r$ is determined only by $\theta$, or that $r$ is a function of $\theta$.

feasible in what sense?

I explained it, didn't I?

Let us call the sequence of values $(\theta,x,y,r)$ feasible if there exists an actual diagram where the parameters in question take these values. For example, if $\theta$ is the angle at $(0,0)$ in the right triangle with vertices $(0,0)$, $(x,0)$ and $(x,y)$ and $r=y/x$, then the sequence $(\pi/6,\sqrt{3}/2,1/2,1/\sqrt{3})$ is feasible, but $(\pi/6,1,0,0)$ is not.

can you explain it in a more comprehensible manner?

I would be glad to if you point out what is not comprehensible in my explanation. That would help me explain myself better next time.

 

[paulmdrdo1]

sorry evegneymakarov. you have to bear with me because i don't really express my thoughts in english fluently thus i couldn't also comprehend english language quickly.

what i want to know is how do you use "feasible" here? "all values x, y, x′, y′, r, r′, if both (θ,x,y,r) and (θ,x′,y′,r′) are feasible"? what do you mean by feasible here? how they are feasible?

thanks!

 

[Evgeny.Makarov]

how do you use "feasible" here? "all values x, y, x′, y′, r, r′, if both (θ,x,y,r) and (θ,x′,y′,r′) are feasible"? what do you mean by feasible here? how they are feasible?

I call $(\theta,x,y,r)$ feasible if there is a drawing with angle $\theta$ and ratio $r$. I mean a drawing of the correct type. You did not describe in the original post what type of drawing you are talking about: namely, what the point $P$ is, what lines form the angle $\theta$ and which segment lengths form the ratio in question. But in any case, I call a pair $(\theta,r)$ (plus any other parameters, such as $x$ and $y$) feasible if there exists a drawing of the type you are considering that has angle $\theta$ and ratio $r$. Thus, $\theta$ determines $r$ if there are no two drawings with the same angle $\theta$, but two different ratios.