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

1. Mar 13, 2007

### Rukawa0320

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

1. The problem statement, all variables and given/known data

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)

2. Relevant 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)

3. 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

2. Mar 13, 2007

### Dick

Can you find a formula for sin(a)-sin(b)? It's pretty standard.

3. Mar 13, 2007

### Rukawa0320

Do you mean sin(A-B) = sinAcosB - cosAsinB ?

4. Mar 13, 2007

### Dick

That's related, but I really do mean sin(a)-sin(b). Look at an extensive table of trig formulae.

Last edited: Mar 13, 2007
5. Mar 13, 2007

### Dick

Or if you want to prove it from scratch, add or subtract your formulas for sin(a+b) and sin(a-b).

6. Mar 13, 2007

### Rukawa0320

oohhh, i found the formula in my exercise book, it was really that standard

thank you for your help, now im doing the (b) part of the exercise about their limits
May I ask for your help if i get stuck?

7. Mar 13, 2007

### Rukawa0320

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

Last edited: Mar 13, 2007
8. Mar 13, 2007

### Dick

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. Mar 13, 2007

### Rukawa0320

I don't really understand what you mean, could you please explain it in details?

10. Mar 13, 2007

### Dick

What's the limit of cos((x+2)/2) as x->2?

11. Mar 13, 2007

### Rukawa0320

isnt it approaches 1?

12. Mar 13, 2007

### Dick

Nope. As x->2, cos((x+2)/2)->cos((2+2)/2)=cos(4/2)=cos(2). Why 1?

13. Mar 13, 2007

### Rukawa0320

But dont you need to take the cosine of two?

14. Mar 13, 2007

### Dick

I could, but it isn't 1. cos(2*pi)=1, not cos(2). Do you agree the answer is cos(2)?

15. Mar 13, 2007

### Rukawa0320

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. Mar 13, 2007

### Dick

It's cos((X0+X0)/2), but that doesn't simplify to what you wrote.

17. Mar 13, 2007

### Rukawa0320

Im getting a lost a little bit, is it actually approaches zero?

18. Mar 13, 2007

### Dick

So am I. Explain how you got cos(X0)/2???

19. Mar 13, 2007

### Rukawa0320

oooh if x-->Xo therefore is it (cos(2*Xo))/2?

20. Mar 13, 2007

### Dick

You are still being sloppy. The '/2' is INSIDE the cosine - not outside.