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Question about how to derive Sine

by alex308
Tags: derive, sine
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alex308
#1
Apr3-09, 03:59 PM
P: 3
So I know that sine of an angle is the side opposite of the angle in a right triangle divided by the hypotenuse. What i want to know is: if i have an angle in a right triangle how do i find the sides of the triangle so i can find Sine.

I spent my study hall trying to figure it out and this is as far as i got. I'm simply using a random angle of 73 to depict my logic.



I figured out that if i could just find the slope of the hypotenuse, i would be able to use point-slope formula to find an equation for the hypotenuse. I could then determine where the hypotenuse intersects the circle which would give me the values of X and Y. I would then be able to calculate the Sine of the angle.

I suppose it all boils down to how do i find the slope of the hypotenuse of a right triangle.
The only one i could figure out was that 45 has a slope of 1 and as theta approaches 90 the slope approaches infinite.

I dont know if this is the original method of deriving Sine but it seems like it could work if i could find the slope of any angle. My intention is to find all the Y values on the circle so that i can plot a sine wave. If anyone can lead me in the right direction with my work or if they could tell me a more logical way of finding sine i would appreciate it.
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dipstik
#2
Apr3-09, 05:01 PM
P: 99
x=r*cos(theta)
y=r*sin(theta)

slope=y/x=tan(theta)=sin(theta)/cos(theta)

does that help, or are you trying to derive this from simpler methods?
jambaugh
#3
Apr3-09, 05:13 PM
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I'm not sure what you mean by "derive" The x and y coordinates on the unit circle are the definition respectively of the cosine and sine functions of the angle.

You can then either define or derive that the slope of the hypotenuse is the tangent of the angle and vis versa that it is the ratio of sine over cosine.

I tell my students to keep straight which is which to remember to keep them in alphabetical order x =cosine, y = sine.

So you can't really "derive" sine and cosine from first principles because that they are the coordinates is the definition i.e. is the "first principles" from which we derive other relations.

Ben Niehoff
#4
Apr3-09, 05:23 PM
Sci Advisor
P: 1,588
Question about how to derive Sine

You could spend days doing this and not get very far. The problem is that most values of the sine function are not constructable numbers.

What you can do is something a bit more useful. Try to use your circle construction to prove the identity:

[tex]\sin(a + b) = \sin a \cos b + \cos a \sin b[/tex]
alex308
#5
Apr3-09, 06:42 PM
P: 3
Quote Quote by jambaugh View Post
I'm not sure what you mean by "derive" The x and y coordinates on the unit circle are the definition respectively of the cosine and sine functions of the angle.

You can then either define or derive that the slope of the hypotenuse is the tangent of the angle and vis versa that it is the ratio of sine over cosine.
I realize that the x and y cordinates are the definitions of cosine and sine. My quarrel is how do i determine the x and y values.

What I meant by derive was how do i come up with the x and y values for this triangle and for all rational degrees around the circle so i can sketch a sine wave.

This is all in an attempt to understand what a calculator is doing when i graph sin(x) or even where it comes up with an answer for sin(73).

If anybody could walk me step be step through finding the sine of 73 without a calculator i would appreciate it.

I know that if someone could just show me how to find the SLOPE of the line that forms the angle between it and then the x axis then i would have the answer to my question.
HallsofIvy
#6
Apr3-09, 07:17 PM
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Thanks
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Unless, you are willing to accept an approximate value, the answer, as you have been told several times now, is that you can't- almost all trig values are NOT "constructible numbers".
alex308
#7
Apr3-09, 07:21 PM
P: 3
Quote Quote by HallsofIvy View Post
Unless, you are willing to accept an approximate value, the answer, as you have been told several times now, is that you can't- almost all trig values are NOT "constructible numbers".
Well that is all well and good but then please explain to me how I could approximate these values.
Ben Niehoff
#8
Apr3-09, 10:28 PM
Sci Advisor
P: 1,588
Do you know calculus? That is the best way to obtain approximate values.
dx
#9
Apr4-09, 03:56 AM
HW Helper
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P: 1,961
You can approximate the values using the power series expansion. There are more sophisticated ways too: http://en.wikipedia.org/wiki/Generat...ometric_tables


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