How did mathematicans define sin?

[primarygun]

How did mathematican define sin A ( A >90 degree) as y/r?
Are there any proofs to sin 150 = sin 30 >?
The proof looks like, sin (180-A)= sin A, how can I get the geometry proof of this ?
Anyone can help me ?

 

[primarygun]

In geometry, how is sin(180-A)=sinA established?
Like sin 120= sin 60
In Geometry, sin 120 seems to be related to nothing before
this established. However, it is said to be proved by
geometry, Could you show me the proof?If I split the angle
into two angles to form two right-angled triangle, that
improves nothing seems sin 120 is a particular value.
Like , if I spilt 120 = 60 + 60, after I had found the
relationship between sin 60 and x-length , y- length ,
radius in the circle, but sin 120 is related to nothing.
Therefore, my effort was useless.My teacher said students
should only memborize it, but I don't agree with him.
Nothing is a gift from the god in maths, I decided to put
my effort forward but rewards little. Anyone helps me ?

 

[cepheid]

In high school (or was it junior high, can't remember now), we were presented with an expanded definition of trigonometric ratios (sine, cosine, and therefore tangent) in terms of a unit circle. Visualise (or better yet, draw) a circle of radius 1 (a unit circle) centred at the origin of an xy coordinate plane (for convenience). Now, your coordinate axes pass through the centre of the circle. We will measure the angles between various radii drawn and the reference axis, which we will define to be the horizontal (the positive x-axis). So, a radius of the circle, a line joining the centre of the circle to some point on it's circumference, forms an angle $\theta$ with the horizontal (the reference axis). Before I continue...have you seen this already in math class?

 

[primarygun]

Yes. Would you like to go on?

 

[cepheid]

Diagram Attached

I drew a diagram to help illustrate the situation, but maybe I should withold comment until it is approved. I'm not sure what constitutes a geometric proof. Nevertheless, the diagram makes it clear that $\sin\theta = \sin(180^o - \theta)$, where $\sin\theta$ is defined as $y/r$ for the coordinate point at which the ray forming the angle theta with the reference axis intersects the circle.

 

[primarygun]

Sorry, but is this the picture?
There is no circle diagram, perhaps something went wrong, could you upload it again?

 

[cepheid]

no...it is still pending approval, as it says. The link will be disabled until it is approved.

 

[cepheid]

Why does it take so long? to get approved?

 

[primarygun]

Would you like to attach onto your own web site?

 

[James R]

Here's an algebraic proof. In general:

$$\sin(A-B) = \sin A \cos B - \cos A \sin B$$

So

$$\sin(180^O -\theta) = \sin 180^O \cos \theta - \cos 180^O \sin \theta = \sin\theta$$

 

[cepheid]

Anyway...the image is there now...plus James' proof using the subtraction formula

 

[primarygun]

Thank you for the nice picture.
I know the meaning of x,y,r.
However, how to prove than sin(180- A)= sin A

 

[cepheid]

It shows it right in the diagram! The two angles theta and 180 - theta clearly have the same y-components...

 

[primarygun]

Sorry for bothering you so much.
sin (180-A) should be the height over hypotheuse in a right-angled triangle.
But there is no right-angled triangle can be formed, how to prove it ?

 

[cepheid]

Sorry for bothering you so much.
sin (180-A) should be the height over hypotheuse in a right-angled triangle.
But there is no right-angled triangle can be formed, how to prove it ?

no...that's the whole point of having the unit circle...like I said before, the defnintions of the trig ratios were expanded so that sin(theta) = y/r. Can you not see in the picture that the line forming an angle of 180 - theta with the reference axis also just forms an angle of theta with the negative x-axis (there's your right triangle)...therefore they have the same y-component...it's obvious...y/r is the same for both.

 

[primarygun]

Isn't the definition of sin A is y/r inside that circle and inside a right-angled triangle.
The angle must be in the triangle, isn't it?

 

[HallsofIvy]

No, No, No.

The fact is that most applications are sine and cosine have nothing to do with triangles. The really important thing about sine and cosine are that they are the "ideal" periodic (repeating) functions and can be used to model periodic phenomena.

The standard definition of sine and cosine is this:

Given the unit circle in some coordinate system, start at (1,0) and measure around the circumference of the circle a distance t (counterclockwise if t>0, clockwise if t<0).
The coordinates of the end point are, by definition, (cos(x), sin(t)).

Notice I said "by definition". You don't need to have any triangles or, indeed, any angles, involved in that at all. You can, of course, draw a right triangle with hypotenuse a radius of the circle so that x and y are the lengths of the legs and determine that x= cos(t)= "opposite side over hypotenuse" and y= sin(t)= "near side over hypotenuse" but those are results of the definition in terms of a circle.

Notice also that t is NOT an angle. It is a length measurement and so its units are given by the coordinate system. Unfortunately, calculators are designed by engineers rather than mathematicians and engineers tend to think of sine and cosine in terms of angles (which is why they talk about things like "phase angle" when there are no angles involved at all!). In order to make the two definitions work together, we have to interpret the "angle" as in RADIANS, not degrees. Except in specific applications where angles are measured in degrees, you should always work in radians.

 

[primarygun]

How to determine an angle given is measured in clockwise or anti-clockwise?

 

[cepheid]

by convention...positive angles are those going anti-clockwise...negative clockwise.