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

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.