# Differentiating sin(x) from first principles

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#1
Do you need to use Taylor's expansions? If not could someone show me how to do it?
Thanks
0
14 years ago
#2
sin(a+b) = sin(a)cos(b) + cos(a)sin(b)

Then remember sin(a)/a -> 1 as a -> 0
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#3
(Original post by 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?
0
14 years ago
#4
Could you use to give and 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.
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14 years ago
#5
(Original post by 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
0
14 years ago
#6
(Original post by 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.
1
14 years ago
#7
(Original post by 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)
0
14 years ago
#8
(Original post by Widowmaker)
dy/dx = sinx/h + cosx - sinx/h
dy/dx = cosx
Hmm?
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#9
(Original post by 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
0
14 years ago
#10
(Original post by Widowmaker)
From first principles:
dy/dx ≈ (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
0
14 years ago
#11
(Original post by 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 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
0
14 years ago
#12
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.
0
14 years ago
#13
(Original post by 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
0
14 years ago
#14
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!
0
14 years ago
#15
And how exactly do you know that (sinx)' = cosx?
0
14 years ago
#16
(Original post by dvs)
And how exactly do you know that (sinx)' = cosx?
lol
I answered for the question lim(sinx/x) not differentiate (original question)
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14 years ago
#17
(Original post by dvs)
And how exactly do you know that (sinx)' = cosx?
Exactly!
0
14 years ago
#18
(Original post by dvs)
And how exactly do you know that (sinx)' = cosx?
Because sin(x) is defined as . 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 , which is the definition of cos(x).
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14 years ago
#19
cool
0
14 years ago
#20
(Original post by Cexy)
Because sin(x) is defined as . 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 , which is the definition of cos(x).
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