Proving slope m of a secant connecting two points of the sine curve

In summary, the slope of the secant connecting the points Po(Xo,Yo) and P(X,Y) of the sine curve is m= sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2).
  • #1
Rukawa0320
17
0
Proving slope "m" of a secant connecting two points of the sine curve

Homework Statement



Write and expression for the slope m of the secant connecting the points Po(Xo,Yo) and P(X,Y) of the sine curve. Use the appropriate trigonometric identity to show that m= sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

Homework Equations



Could somebody give a hint how to do the second proof? m= sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

Basically what i need to prove is that:
(Sin(Xo)-Sin(X)) / Xo-X = sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

The Attempt at a Solution



I could manage to do the first part of the exercise, as far as now I got a formula which is the following:

m= (Yo-Y)/(Xo-x) = (Sin(Xo)-Sin(X)) / Xo-X
and the expression (Sin(Xo)-Sin(X)) / Xo-X is equal to m
 
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  • #2
Can you find a formula for sin(a)-sin(b)? It's pretty standard.
 
  • #3
Do you mean sin(A-B) = sinAcosB - cosAsinB ?
 
  • #4
That's related, but I really do mean sin(a)-sin(b). Look at an extensive table of trig formulae.
 
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  • #5
Or if you want to prove it from scratch, add or subtract your formulas for sin(a+b) and sin(a-b).
 
  • #6
oohhh, i found the formula in my exercise book, it was really that standard

thank you for your help, now I am doing the (b) part of the exercise about their limits
May I ask for your help if i get stuck?
 
  • #7
Im doing the (b) part of the exercise, which ask about what do they approach as X approaches Xo (their limits)

1)sin((X-Xo)/2)/((X-Xo)/2)
2)cos ((X+Xo)/2)
3)m(=sin((X-Xo)/2)/((X-Xo)/2) * cos ((X+Xo)/2)

Note: Xo an X are points on the cos,sin curve which is not bigger than pi/2

I was able to solve the first one, I got as X approaches Xo it will approach 1.(I think I did right, but please tell me if it's wrong)
I got stuck in the second one, could you guys give me some hint to solve the 2.?
In the thrid I guess all I need to do is to multiply the two limits (1. and 2.) and that's it.

Any helps appreciated
 
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  • #8
The second one is MUCH easier than the first one. X -> X0. No fancy limits to take really. The answer is a function of X0 - it's not a constant like the first one, if that is what is throwing you off.
 
  • #9
I don't really understand what you mean, could you please explain it in details?
 
  • #10
What's the limit of cos((x+2)/2) as x->2?
 
  • #11
isnt it approaches 1?
 
  • #12
Nope. As x->2, cos((x+2)/2)->cos((2+2)/2)=cos(4/2)=cos(2). Why 1?
 
  • #13
But don't you need to take the cosine of two?
 
  • #14
I could, but it isn't 1. cos(2*pi)=1, not cos(2). Do you agree the answer is cos(2)?
 
  • #15
I see what you mean, that (2) is in radian and not degrees, thus the answer to the original question is (cos(Xo))/2 right?
 
  • #16
It's cos((X0+X0)/2), but that doesn't simplify to what you wrote.
 
  • #17
Im getting a lost a little bit, is it actually approaches zero?
 
  • #18
So am I. Explain how you got cos(X0)/2?
 
  • #19
oooh if x-->Xo therefore is it (cos(2*Xo))/2?
 
  • #20
You are still being sloppy. The '/2' is INSIDE the cosine - not outside.
 
  • #21
than its simply cosXo
if I multiply it with the limit of 1st one :1 than the limit of m approaches cosXo?
 
  • #22
Yes. You just differentiated sin(x).
 
  • #23
wow, it feels great when you finally solve something which you work on for a long time

thanks for your help!
 

What is the formula for finding the slope of a secant on the sine curve?

The formula for finding the slope (m) of a secant connecting two points (x1,y1) and (x2,y2) on the sine curve is: m = (y2-y1)/(x2-x1).

How do you prove the slope of a secant on the sine curve?

To prove the slope of a secant on the sine curve, you need to find the slope of the line connecting the two points on the curve. This can be done by using the formula m = (y2-y1)/(x2-x1). Then, you can compare this slope to the slope of the sine curve at any given point to verify that they are equal.

What are the steps for finding the slope of a secant on the sine curve?

The steps for finding the slope of a secant on the sine curve are:

  1. Identify the two points on the sine curve.
  2. Plug the coordinates of the two points into the formula m = (y2-y1)/(x2-x1).
  3. Simplify the equation to find the slope.
  4. Compare the slope to the slope of the sine curve at any given point to verify that they are equal.

Can the slope of a secant on the sine curve be negative?

Yes, the slope of a secant on the sine curve can be negative. This means that the secant line is decreasing as you move from left to right on the curve.

What is the significance of finding the slope of a secant on the sine curve?

The slope of a secant on the sine curve is significant because it represents the average rate of change of the sine function between two points. It can also be used to approximate the slope of the tangent line at a specific point on the curve.

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