Can't Get Formula Result - Excel Radian Issue

[ISX]

I can't get this formula to come out. I have done it on excel step by step over and over and it gets close, but always a little under. As in if the chord angle is 45 and I do all the calculations on the other side, it calculates to 43.85. They should both be 45 so I don't know what is going on. I figured out excel uses radians so I have it convert everything over first, then back to degrees.

 

[Unrest]

The equation isn't right.
The right hand side is the length of a chord on a unit circle
The left hand side (theta) is the length of the arc that subtends an angle of theta on a unit circle.

When theta is very small they're approximately equal, but the higher theta, the more the chord separates from the edge of the circle, so the lengths become more different.

There is an exact solution at theta=0.

 

[ISX]

Hmmm. Well I keep seeing this formula over and over when looking up the history of sine. Here is where I got this formula, it talks about how sine is a half chord. So if the equation isn't right, why do so many sites have it wrong? http://en.wikipedia.org/wiki/History_of_trigonometry#Greek_mathematics

 

[ISX]

I also don't understand how the sine of an angle is half the chord of twice the angle. That would mean the sine of theta would = the opposite side, so why do we also have a formula for opposite over hypotenuse to find sine of theta? Seems redundant.

 

[Char. Limit]

Define the length of the chord as crd(theta). Then if you draw a triangle with crd(theta) as the hypotenuse, you find that the other two sides of the triangle are sin(theta) and 1-cos(theta). Thus...

crd(\theta) = \sqrt{(1-cos(\theta))^2 + sin^2(\theta)} = \sqrt{2-2cos(\theta)} = 2 \sqrt{\frac{1-cos(\theta)}{2}}

Then, by the half-angle formula for the sine...

2\sqrt{\frac{1-cos(\theta)}{2}} = 2 sin\left(\frac{\theta}{2}\right)

Thus, crd(\theta)= 2 sin\left(\frac{\theta}{2}\right)

Does that help?

 

[ISX]

Helps a little but is over my head. I am trying to understand where sine was derived and how they came up with it. The half chord thing has me confused since it doesn't calculate right.

When you say chord (theta) as the hypotenuse, I thought theta was the angle, therefore it can't be the hypotenuse since that is a side. Now you see how confused I am lol.

 

[tiny-tim]

Hi ISX!

(have a theta: θ )

I am trying to understand where sine was derived and how they came up with it.

If A and B are on a circle of centre O and radius 1, then AB is the chord, and AOB is the angle θ.

Draw OD perpendicular to AB, then AD = BD = AB/2, and DOA = DOB = θ/2, and AO = BO = 1,

so sin(θ/2) = AD/AO = (1/2 chordAB)/1,

ie chordAB = 2sin(θ/2)

 

[Goongyae]

>I am trying to understand where sine was derived and how they came up with it.

I believe it was originally an empirical measurement based on triangles.

But there are now plenty of suitable, precise choices not relying on triangles. For example:
1. sin(x) = (exp(ix)-exp(-ix))/(2i), where i = sqrt(-1)
2. The infinite polynomial, sin(x) = x - x^3/3! + x^5/5! - x^7/7! + etc.
3. sin(x) = pi / (gamma(x/pi) gamma(1-x/pi))
4. The infinite product, sin(x) = x(1-x/pi^2)(1-x/(4pi^2))(1-x/(9pi^2))*...

Your hand calculator or computer processor most likely uses #2 or perhaps a rational function approximation.

 

[ISX]

Hi ISX!

(have a theta: θ )


If A and B are on a circle of centre O and radius 1, then AB is the chord, and AOB is the angle θ.

Draw OD perpendicular to AB, then AD = BD = AB/2, and DOA = DOB = θ/2, and AO = BO = 1,

so sin(θ/2) = AD/AO = (1/2 chordAB)/1,

ie chordAB = 2sin(θ/2)

This made some sense until I actually calculated it, then I am lost again. I have been using the 3/4/5 triangle since it is simple. So using "sin(θ/2) = AD/AO = (1/2 chordAB)/1", I can say that AD is 3, AO is 4. Well 3/4 = 0.75. Now 1/2 chord AB means the chord would be twice of AD or 6, half that would be 3. I am not seeing how 3 = 0.75. That is where I get all the confusion, there is something I am missing but I can't understand it, I don't see how things equal each other when they don't. Thanks for the help so far!

 

[Goongyae]

> I am not seeing how 3 = 0.75.

You're confusing two things.

"Chord" is a word meaning a certain part of a triangle that touches a circle. In your case the chord has phyiscal length AB and so it's 3 inches, or 3 cm, or if you weren't worrying about units, it's just 3 whatevers.

"Chord" also has a different meaning: it's the name of a mathematical function that returns a dimensionless number, i.e. a ratio. For you, the chord function yields 0.75. This function, like other trig functions, is basically designed to support a circle of unit radius. Since your circle has a radius of 4 inches though, you need to take that into account: 0.75 x 4" yields 3", the length of your physical chord.

A similar confusion exists for a number of trigonometric terms. All of these are both mathematical functions and parts of physical diagrams: sine, cosine, tangent, secant, exsecant, and chord. In practice, I think most people don't run into this confusion because they usually just use these words to refer to the mathematical functions and almost never to actual parts of physical diagrams, with physical sizes.

If you insist on using these words for both mathematical functions, and for parts of diagrams, you will continue to confuse yourself. It will help at least if you think about the parts of diagrams as having units such as inches, or centimeters.

By the way, these functions are very archaic and seldom used in mathematics: chord, exsecant, versine, haversine, coversine, cohaversine, hacoversine, etc. Also, apothem and sagittus are parts of triangles/diagrams but are never used to refer to functions. You won't find ANY of these on a modern calculator.

You're better off getting used to the commonly used functions and expressing in terms of them: sine, cosine, secant, tangent, cotangent, cosecant.

 

[tiny-tim]

Hi ISX!

(just got up :zzz: …)

… I can say that AD is 3, AO is 4.

No, AO, the radius of the circle, is the hypotenuse of the triangle, 5.

(sin = opp/hyp)

 

[ISX]

Alright I am really close to understanding it. Now I just want to see all this in action so I can actually see how you guys calculate it. I drew the circle and triangle and everything.

So can you guys show me how this formula "sin(θ/2) = AD/AO = (1/2 chordAB)/1" works out, showing how each part before the equal sign is calculated. Then how this one is done "chordAB = 2sin(θ/2)".

Here's the triangle to go by.

Thanks for helping me out with this.

 

[tiny-tim]

Hi ISX!

You can read it straight off the diagram …

θ = AOB,

so chordAB = 2AD = 2sinAOD = 2sin(θ/2) …

what is worrying you about that? ​

 

[ISX]

I can't figure out how it calculates out, I need to see it done on a triangle then I can figure it all out. ChordAB is 6, 2AD is 6, but the sin of AOD is 0.6 * 2 is 1.2.

 

[tiny-tim]

oh, the radius is 5 …

the radius in the original question was 1 …

ok, divide by 5 …

sinAOD = opp/hyp = 3/5 = 0.6

chordAB = AB/5 = 1.2 = 2sinAOD ​

 

[ISX]

A post before this last one you said "so chordAB = 2AD = 2sinAOD = 2sin(θ/2)" and here you say "chordAB = AB/5 = 1.2 = 2sinAOD". From what I can gather, all of that is all equal. Which means 2AD equals AB/5 or 1.2. Well how does 2AD = 1.2 when line AD is 3, so 2AD is 6?

 

[Mentallic]

2AD is 6, but the sin of AOD is 0.6 * 2 is 1.2.

sin(AOD)=3/5=0.6 yes, but that doesn't tell us what the opposite length is.

In any right triangle, \sin\theta=\frac{opp}{hyp} so in order to find the length you're looking for (which is the opposite length), you need to multiply through by the hypotenuse length, so we have opp=hyp\cdot\sin\theta which will give you the value 3 which you're looking for.

By the way, sin(AOD) is not a length, it's a ratio between the opposite and hypotenuse sides. Since it's equal to 0.6, this means the opposite length to hypotenuse length is in a ratio of 0.6:1, or 3:5, or 3000:5000 etc. The sine of an angle doesn't tell us anything about the size of the triangle, only the ratios of the sides.

 

[zgozvrm]

As Unrest stated, that equation is in reference to the unit circle (one with a radius of 1).
For the general circle, the equation becomes:

\text{chord} ~ \theta = 2r \sin \frac{\theta}{2}

This is basically what Mentallic was pointing out.