Trigonometry

[Primarygun]

How did mathematican define sin A ( A >90 degree) as [MATH]y/r[/MATH]?

Are there any proofs to [MATH]sin 150 = sin 30 [/MATH]>?

[Primarygun]

How did mathematican define sin A ( A >90 degree) as [MATH]y/r[/MATH]?

Are there any proofs to [MATH]sin 150 = sin 30 [/MATH]>?

[premjan]

I'm thinking that the unit circle was probably used to define all the trig ratios. If you draw a unit circle and then look at the radius line that makes an angle starting from 0deg to 360 deg turning ccw, you could easily prove the various trig identities.

[premjan]

I'm thinking that the unit circle was probably used to define all the trig ratios. If you draw a unit circle and then look at the radius line that makes an angle starting from 0deg to 360 deg turning ccw, you could easily prove the various trig identities.

[Primarygun]

The proof looks like, sin (180-A)= sin A, how can I get the geometry proof of this ?

Anyone can help me ?

[Primarygun]

The proof looks like, sin (180-A)= sin A, how can I get the geometry proof of this ?

Anyone can help me ?

[premjan]

just set r=1, and draw the diagram. it is rather obvious and I am uncertain what steps a formal geometric proof could require. Basically, the angle 180-A is obtained by reflection of the angle A around the y-axis, so pretty much by that definition, the y-coordinate remains the same, and r is not going to change since it is a circle.

[premjan]

just set r=1, and draw the diagram. it is rather obvious and I am uncertain what steps a formal geometric proof could require. Basically, the angle 180-A is obtained by reflection of the angle A around the y-axis, so pretty much by that definition, the y-coordinate remains the same, and r is not going to change since it is a circle.

[Primarygun]

Yes I know.

By the diagram, how can we prove sin A ( 180>A>90 ) is y/r ?

The algebra can prove this but the algebra is proved by this geometry, that's say if the geometry proof is wrong , algebra of the proof is wrong indeed. Can you show me the proof ? I can't find it in my book or web site.

[Primarygun]

Yes I know.

By the diagram, how can we prove sin A ( 180>A>90 ) is y/r ?

The algebra can prove this but the algebra is proved by this geometry, that's say if the geometry proof is wrong , algebra of the proof is wrong indeed. Can you show me the proof ? I can't find it in my book or web site.

[premjan]

what is the definition of sin(theta) that you use? Assuming that it is opposite/hypotenuse, how would you extend this to the case where theta > 90? for that matter, how do you reconcile the fact that at 90deg, there is effectively no right triangle at all?

[premjan]

what is the definition of sin(theta) that you use? Assuming that it is opposite/hypotenuse, how would you extend this to the case where theta > 90? for that matter, how do you reconcile the fact that at 90deg, there is effectively no right triangle at all?

[Primarygun]

Yes. I don't know how to reconcile.

Therefore, that is my question asked.

[Primarygun]

Yes. I don't know how to reconcile.

Therefore, that is my question asked.

[maverick88]

The proof looks like' date=' sin (180-A)= sin A, how can I get the geometry proof of this ?

Anyone can help me ?[/quote']

 

ain(180-a) = sina because sinus is uneven function, to say sina=-sin(-a).

and because of sin beeng periodic (360 deg.) u can see that its values are same at 0 180 360 deg.

I hope i got what i mean

[maverick88]

The proof looks like' date=' sin (180-A)= sin A, how can I get the geometry proof of this ?

Anyone can help me ?[/quote']

 

ain(180-a) = sina because sinus is uneven function, to say sina=-sin(-a).

and because of sin beeng periodic (360 deg.) u can see that its values are same at 0 180 360 deg.

I hope i got what i mean

[Primarygun]

Yes.

Using a calculator can find the value of sin (180-a)=sin A

But this definition is proved inside the geometry circle first.

Do you know the proof?

[Primarygun]

Yes.

Using a calculator can find the value of sin (180-a)=sin A

But this definition is proved inside the geometry circle first.

Do you know the proof?

[premjan]

you should use the following definition of sin, then it is very easy to reconcile.

 

draw a unit circle centered at (0,0). let the angle theta be the angle that a radius of this circle makes with the +x axis. Let sin theta be the ratio y/r where (x,y) is the point at which the given radius intersects the circle (there is only one such point).

 

Then use the reflection argument which I gave you earlier.

[premjan]

you should use the following definition of sin, then it is very easy to reconcile.

 

draw a unit circle centered at (0,0). let the angle theta be the angle that a radius of this circle makes with the +x axis. Let sin theta be the ratio y/r where (x,y) is the point at which the given radius intersects the circle (there is only one such point).

 

Then use the reflection argument which I gave you earlier.

[maverick88]

Yes.

Using a calculator can find the value of sin (180-a)=sin A

But this definition is proved inside the geometry circle first.

Do you know the proof?

 

In order to understand the proof you have to uderstand the features of the function Sin: In every 180 degrees there are two equal values of sin (sin = the Y value of a tangle(any tabgle you choose)) besides 1 and -1. If so you can see that if a distance of two angles frim the Y-axis is equal-->their sin is equal

[maverick88]

Yes.

Using a calculator can find the value of sin (180-a)=sin A

But this definition is proved inside the geometry circle first.

Do you know the proof?

 

In order to understand the proof you have to uderstand the features of the function Sin: In every 180 degrees there are two equal values of sin (sin = the Y value of a tangle(any tabgle you choose)) besides 1 and -1. If so you can see that if a distance of two angles frim the Y-axis is equal-->their sin is equal

[premjan]

I think this last definition obviates the need for a proof (more of proof by examination than anything).

[premjan]

I think this last definition obviates the need for a proof (more of proof by examination than anything).

[psi20]

Another way of looking at it is through reference angles.

Here's another way of looking at it.

I use the letters a and b as angle names.

 

You have your unit circle. Draw a point on (0,1). Draw a point on (cos a, sin a), preferably a<90.

Draw a second point at (cos b, sin b) where a + b = 180.

 

Now, if a + b = 180 and a<90, then b - 90 = 90 - a.

Therefore, (cos a, sin a) and (cos b, sin b) are equidistant from (0,1)

 

Well,

sqrt(cos^2 y... you'll have to do this on paper. It'll get too confusing at this rate.

 

Use the distance formula.

Then square both sides to get rid of the square root.

Expand.

Use the idea that cos^2 X + sin^2 X = 1

The two 1's on each side cancel out.

Divide both sides by -2.

You're left with sin a = sin b

 

Since a + b = 180, b = 180 - a.

sin a = sin (180-a)