Sinus domain: y = sin(alpha) What is the domain of y is between -1 to 1?

[shahar]

y = sin(alpha)
What is the domain of y is between -1 to 1?

 

[Deleted member 4993]

y = sin(alpha)
What is the domain of y is between -1 to 1?

I am assuming you are assigning "y" as a dependant variable. In that it is the "range" of "y" you are seeking.

If you had written:

\(\displaystyle \alpha \) = sin^-1(y)

Then you could talk about the domain of "y" - and that would be -1 and 1 (inclusive).

 

[shahar]

I am assuming you are assigning "y" as a dependant variable. In that it is the "range" of "y" you are seeking.

If you had written:

\(\displaystyle \alpha \) = sin^-1(y)

Then you could talk about the domain of "y" - and that would be -1 and 1 (inclusive).

Why the domain of y would be -1 and -1?
The underlined text in your reply.
What the cause of it?
How can I Proof it?

 

[sinx]

Why the domain of y would be -1 and -1?
The underlined text in your reply.
What the cause of it?
How can I Proof it?

because the sin of an angle is never larger than 1 (or -1).
[@ 90⁰ and 270⁰​]

 

[JeffM]

"Domain" is the set of all possible values of the independent variable or variables for which the function is defined.

"Range" is the set of of all possible values that the dependent value may take.

When you write \(\displaystyle y = sin(x)\), you are implying that y is the dependent variable.

Therefore, asking what is the domain of y makes absolutely no sense as Subhotosh Khan has already explained.

If instead you are asking what is the range of y, then sin(x) never exceeds 1 nor is exceeded by - 1 as was explained by sinx.

\(\displaystyle \therefore -\ 1 \le sin(x) \le 1 \text { and } y = sin(x) \implies -\ 1 \le y \le 1.\)

 

[shahar]

\(\displaystyle \therefore -\ 1 \le sin(x) \le 1 \text { and } y = sin(x) \implies -\ 1 \le y \le 1.\)

So there is no geometric/trigonometric explanation?
I don't understand.
Can you expalian why yes or why not?

 

[Steven G]

because the sin of an angle is never larger than 1 (or -1).
[@ 90⁰ and 270⁰​]

Actually the sin of an angle is always greater than or equal to -1

 

[Steven G]

So there is no geometric/trigonometric explanation?
I don't understand.
Can you expalian why yes or why not?

Let's just talk now about why sinx can be between 0 and 1. Consider a right angle with the base leg having a length of 1 (any other number will work). Let the angle between the base leg and the hypotenuse be x. Lets call the opposite side y. If y has length of 0, then the angle x = 0^o. So the sin(0) = opp/hyp = 0/1 = 0. I say that the hypotenuse is 1 since if a right angle has a 0^o angle, then the hypotenuse lays right on top of the base which has length = 1). Now consider what happens if y is extremely large (and the base is still 1). What happens is the opp and hyp are about the same length. In fact if we keep letting y get larger and other, then sin(x) gets closer and closer to 1. In fact the sin (90^o) =1. Since the sin of some angles can be negative, we also have that the sin(x) can be between -1 and 0. Combining we have sin(x) is between -1 and 1

 

[Deleted member 4993]

So there is no geometric/trigonometric explanation?
I don't understand.
Can you expalian why yes or why not?

- geometric/trigonometric explanation for what?
Please state (may be restate your question) regarding explanation.

 

[JeffM]

So there is no geometric/trigonometric explanation?
I don't understand.
Can you expalian why yes or why not?

If you define the sine function in terms of the ratio of the length of the side opposite to an angle of \(\displaystyle \theta\) in a right triangle over the length of the triangle's hypotenuse, you get the following definition

\(\displaystyle sin( \theta ) = \dfrac{ \omega }{ \lambda }, \text { where}\)

\(\displaystyle \theta = \text { the numeric measure of an angle in a right triangle;}\)

\(\displaystyle \omega = \text { the length of the side opposite the angle in question, and}\)

\(\displaystyle \lambda = \text { the length of the right triangle's hypotenuse.}\).

A length is a positive number so

\(\displaystyle \omega > 0 \text { and } \lambda > 0 \implies sin( \theta ) = \dfrac{\omega}{\lambda} > 0.\)

The longest side in a right triangle is the hypotenuse, which is also the side opposite the right angle. So

\(\displaystyle sin(90^o) = \dfrac{\lambda}{\lambda} = 1.\)

Thus, if you define the sine function on the basis of the geometry of right triangles, you get

\(\displaystyle 0 < sin( \theta ) \le 1\)

because the domain of \(\displaystyle \theta\) is only \(\displaystyle (0,\ 90^o).\)

But the modern definitions of the trigonometric functions are not tied to Euclidean geometry.

https://en.m.wikipedia.org/wiki/Trigonometric_functions

If, as is the modern approach, you define the trigonometric functions in terms of the unit circle on a Cartesian plane or in terms of infinite series, you get

\(\displaystyle -\ 1 \le sin(\theta) \le 1.\)

It is far easier to grasp why that is so in terms of the unit circle.

EDIT: I have changed the argument of the sine function to \(\displaystyle \theta\), as is consistent with former practice, because it will prevent confusion if the OP moves on to the definition of the sine in terms of the unit circle.

 

[shahar]

If you define the sine function in terms of the ratio of the length of the side opposite to an angle of \(\displaystyle \theta\) in a right triangle over the length of the triangle's hypotenuse, you get the following definition

\(\displaystyle sin( \theta ) = \dfrac{ \omega }{ \lambda }, \text { where}\)

\(\displaystyle \theta = \text { the numeric measure of an angle in a right triangle;}\)

\(\displaystyle \omega = \text { the length of the side opposite the angle in question, and}\)

\(\displaystyle \lambda = \text { the length of the right triangle's hypotenuse.}\).

A length is a positive number so

\(\displaystyle \omega > 0 \text { and } \lambda > 0 \implies sin( \theta ) = \dfrac{\omega}{\lambda} > 0.\)

The longest side in a right triangle is the hypotenuse, which is also the side opposite the right angle. So

\(\displaystyle sin(90^o) = \dfrac{\lambda}{\lambda} = 1.\)

Thus, if you define the sine function on the basis of the geometry of right triangles, you get

\(\displaystyle 0 < sin( \theta ) \le 1\)

because the domain of \(\displaystyle \theta\) is only \(\displaystyle (0,\ 90^o).\)

2:
But the modern definitions of the trigonometric functions are not tied to Euclidean geometry.

https://en.m.wikipedia.org/wiki/Trigonometric_functions

If, as is the modern approach, you define the trigonometric functions in terms of the unit circle on a Cartesian plane or in terms of infinite series, you get

\(\displaystyle -\ 1 \le sin(\theta) \le 1.\)

It is far easier to grasp why that is so in terms of the unit circle.

EDIT: I have changed the argument of the sine function to \(\displaystyle \theta\), as is consistent with former practice, because it will prevent confusion if the OP moves on to the 1: definition of the sine in terms of the unit circle.

[The bold and underline text in the quoted reply above:]
(1) What is the definition of sine in terms of the unit circle?
(1.b) Why you mention it?
(1.c) Why the difference between the two notifications?


(2) Can you explain if the Euclidean geometry can be subset of other geometry?

 

[Deleted member 4993]

[The bold and underline text in the quoted reply above:]
(1) What is the definition of sine in terms of the unit circle? please refer to: https://courses.lumenlearning.com/b.../trigonometric-functions-and-the-unit-circle/
(1.b) Why you mention it?
(1.c) Why the difference between the two notifications?


(2) Can you explain if the Euclidean geometry can be subset of other geometry?

We made observations about the angles and sides of right triangles, but these observations were limited to angles with measures less than 90^∘. Using the unit circle, we are able to apply trigonometric functions to angles greater than 90^∘.

 

[shahar]

We made observations about the angles and sides of right triangles, but these observations were limited to angles with measures less than 90^∘. Using the unit circle, we are able to apply trigonometric functions to angles greater than 90^∘.

Still, I don't understand this:
Correct me, but you said you change the angle from some sign to theta. Why?
If you say something else, write it in another words.
Thanks!

 

[JeffM]

[The bold and underline text in the quoted reply above:]
(1) What is the definition of sine in terms of the unit circle?

Did you bother to read the link I gave in my first post?

(1.b) Why you mention it?
(1.c) Why the difference between the two notifications?

Because (1) the definition based on the unit circle leads to non-positive values of the sine function whereas the definition based on the right triangle leads only to positive values, and (2) the definition based on the unit circle leads to the sine function that is primarily used in modern mathematics and science.

(2) Can you explain if the Euclidean geometry can be subset of other geometry?

The Euclidean plane is the foundation for the Cartesian and Argand planes.

Eucldean geometry is an element in the set of Riemannian geometries, not a subset.

 

[JeffM]

Still, I don't understand this:
Correct me, but you said you change the angle from some sign to theta. Why?
If you say something else, write it in another words.
Thanks!

No, I recommended using \(\displaystyle sin( \theta )\) rather than \(\displaystyle sin(x).\)

The reason is that, using the standard labels for the horizontal and vertical axes of the Cartesian plane, defining the sine and cosine in terms of the unit circle leads to the following formulas:

\(\displaystyle sin( \theta ) = y \text { and } cos ( \theta ) = x.\)

 

[Deleted member 4993]

Still, I don't understand this:
Correct me, but you said you change the angle from some sign to theta. Why?
If you say something else, write it in another words.
Thanks!

Did you study the website I referred to? That website is:

https://courses.lumenlearning.com/bo...e-unit-circle/

Does that answer the question about trigonometric definitions through unit circle?

 

[HallsofIvy]

[The bold and underline text in the quoted reply above:]
(1) What is the definition of sine in terms of the unit circle?

In a Cartesian coordinate system, construct a "unit circle", the circle with center at (0, 0) and radius 1. For any point, (x, y), on that circle, \(\displaystyle x^2+ y^2= 1\). Starting at the point (1, 0) on the circle, measure a distance "t" around the circumference. "sin(t)" is defined as the y coordinate of the endpoint and "cos(t)" is the x coordinate.

(1.b) Why you mention it?

Probably because while you ask about "sin(x)" you say nothing about its definition or what you know about sin(x).

(1.c) Why the difference between the two notifications?

The "trig" definition requires that the argument be between 0 and 90 degrees (0 and \(\displaystyle \pi/2\) radians). The "circle" does not- the argument can be any real (or with an extended definition, complex) number so more amenable to use in mathematical models. Also using the "trig" definition you must state whether you are measuring the angle in degrees or radians. In the "circle" definition there are no units. What "distance t around the unit circle" means is determined when you construct the original Cartesian coordinate system and unit circle.

(2) Can you explain if the Euclidean geometry can be subset of other geometry?

In a Cartesian coordinate system we have a natural notion of "distance". Other geometries may have no defined "distance" but we can restrict to Euclidean geometry by defining an appropriate distance.

 

[shahar]

12

Another question:
When I use the therefore sign:
\(\displaystyle \therefore\)
And when I use the implies sign:
\(\displaystyle \implies\)

Can each one replace each other?

 

[JeffM]

Another question:
When I use the therefore sign:
\(\displaystyle \therefore\)
And when I use the implies sign:
\(\displaystyle \implies\)

Can each one replace each other?

Yes.

 

[Otis]

… What is the domain of y is between -1 to 1?

This is not proper English. Is there a typo, maybe?