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

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

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

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

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

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

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

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:
Dick
Homework Helper
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.

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

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

isnt it approaches 1?

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

But dont you need to take the cosine of two?

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

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?

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

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

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

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

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

than its simply cosXo
if I multiply it with the limit of 1st one :1 than the limit of m approaches cosXo?

Dick