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sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

Then remember sin(a)/a -> 1 as a -> 0
Reply 2
AlphaNumeric
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

Then remember sin(a)/a -> 1 as a -> 0

Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.
I can sort of see from the graph of y=sinx that it (looks like it) tends to y=x but is there a better reason?
Reply 3
Could you use eix=cosx+isinxe^{ix} = \cos x + i\sin x to give sinx=eixeix2i\sin x = \frac{e^{ix}-e^{-ix}}{2i} and cosx=eix+eix2\cos x = \frac{e^{ix}+e^{-ix}}{2} and note that differentiating this expression for sin x gives the expression for cos x, or does this formula presuppose Taylor's expansion (the only way I can think of proving it does)?

Edit: the wierd s-like thingies in the exponents in the formulae are actually x's which have been mutilated by being rendered at too low a resolution. :biggrin:
e-unit
Do you need to use Taylor's expansions? If not could someone show me how to do it?
Thanks


From first principles:
dy/dx ≈ lim (f(x+h)-f(x))/(h+x-x) as h-->0

So for sinx
dy/dx = limh-->0(sin(x+h)-sinx)/h
dy/dx = limh-->0(sinxcosh+sinhcosx-sinx)/h

as h --> 0 sinh/h --> 1 and cosh ---> 1
Therefore, for small values of h
sinh ≈ h

dy/dx = (sinx x 1 + hcosx - sinx)/h
dy/dx = (sinx+hcosx-sinx)h
dy/dx = hcosx/h
dy/dx = cosx
Reply 5
e-unit
Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.
I can sort of see from the graph of y=sinx that it (looks like it) tends to y=x but is there a better reason?

You can prove it geometrically by drawing the unit circle and considering a sector with angle 0<x<pi/2, then comparing the areas of the inscribed triangle, the sector and the circumscribed triangle. That would give you the inequality:
cos(x)sin(x)/2 <= x/2 <= tan(x)/2

Which can be simplified to:
cos(x) <= sin(x)/x <= 1/cos(x)

And letting x -> 0 we find sin(x)/x -> 1.
Reply 6
e-unit
Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.

It's a bit complicated proof, using geometry as I studied before. But maybe there's another way (I don't mention about Taylor series or L'Hopital as it might not be taught in your degree)
Reply 7
Widowmaker
dy/dx = sinx/h + cosx - sinx/h
dy/dx = cosx

Hmm?
Reply 8
BCHL85
It's a bit complicated proof, using geometry as I studied before. But maybe there's another way (I don't mention about Taylor series or L'Hopital as it might not be taught in your degree)


Or A-Level course :biggrin:
Reply 9
Widowmaker
From first principles:
dy/dx &#8776; (f(x+h)-f(x))/(h+x-x) lim h --> 0

So for sinx
dy/dx = (sin(x+h)-sinx)/h lim h-->0
dy/dx = (sinxcosh+sinhcosx-sinx)/h lim h-->0
dy/dx = sinxcosh/h + sinhcosx/h - sinx/h

as h --> 0 sinh/h --> 1 and cosh ---> 1

dy/dx = sinx/h + cosx - sinx/h
dy/dx = cosx

That may look like it works, but you can't just let h->0 then leave terms like sin(x)/h in there.

You have to reconsider this:
dy/dx = lim [sin(x)cos(h)]/h + [sin(h)cos(x)]/h - sin(x)/h

Perhaps re-writing it as this and using an identity might help:
lim [sin(x)(cos(h)-1)]/h + lim [sin(h)cos(x)]/h
e-unit
Do you need to use Taylor's expansions? If not could someone show me how to do it?
Thanks


From first principles:
dy/dx &#8776; lim (f(x+h)-f(x))/(h+x-x) as h-->0

So for sinx
dy/dx = limh-->0(sin(x+h)-sinx)/h
dy/dx = limh-->0(sinxcosh+sinhcosx-sinx)/h

as h --> 0 sinh/h --> 1 and cosh ---> 1
Therefore, for small values of h
sinh &#8776; h so replace sinh with h and cosh with 1.

dy/dx = (sinx x 1 + hcosx - sinx)/h
dy/dx = (sinx+hcosx-sinx)h
dy/dx = hcosx/h
dy/dx = cosx
Reply 11
Perhaps the best way is to use geometry to find suitable functions for the comparison test.
Using L'Hopitals rule would create a circular argument.
Reply 12
Gaz031
Perhaps the best way is to use geometry to find suitable functions for the comparison test.
Using L'Hopitals rule would create a circular argument.

No, you know (sinx)' = cosx, and (x)' = 1
=> lim(sinx/x) = lim(cosx/1)
Then when x -> 0, cosx -> cos0 = 1
Reply 13
yar thats L'hospital's territory, if you have a limit with indeterminate form 0/0 or inf./inf., and both sin x and x go to zero, but dx/dx = 1 so there is no circular argument in this path as the denominator vanishes! we win!
Reply 14
And how exactly do you know that (sinx)' = cosx? :smile:
Reply 15
dvs
And how exactly do you know that (sinx)' = cosx? :smile:

lol
I answered for the question lim(sinx/x) not differentiate (original question)
Reply 16
dvs
And how exactly do you know that (sinx)' = cosx? :smile:

Exactly!
Reply 17
dvs
And how exactly do you know that (sinx)' = cosx? :smile:

Because sin(x) is defined as Σn=0(1)nx2n+1(2n+1)!\Sigma_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. Theorems about power series tell us that this has infinite radius of convergence, and thus its derivative has infinite radius of convergence. Its derivative is Σn=0(1)nx2n(2n)!\Sigma_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}, which is the definition of cos(x).
Reply 18
cool :smile:
Reply 19
Cexy
Because sin(x) is defined as Σn=0(1)nx2n+1(2n+1)!\Sigma_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. Theorems about power series tell us that this has infinite radius of convergence, and thus its derivative has infinite radius of convergence. Its derivative is Σn=0(1)nx2n(2n)!\Sigma_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}, which is the definition of cos(x).

:rolleyes: