# Differentiating sin(x) from first principles

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Do you need to use Taylor's expansions? If not could someone show me how to do it?

Thanks

Thanks

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(Original post by

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

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

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

Then remember 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?

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#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.

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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#5

(Original post by

Do you need to use Taylor's expansions? If not could someone show me how to do it?

Thanks

**e-unit**)Do you need to use Taylor's expansions? If not could someone show me how to do it?

Thanks

dy/dx ≈ lim (f(x+h)-f(x))/(h+x-x) as h-->0

So for sinx

dy/dx = lim

_{h-->0}(sin(x+h)-sinx)/h

dy/dx = lim

_{h-->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

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#6

(Original post by

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?

**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?

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.

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#7

(Original post by

Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.

**e-unit**)Yeah that's what I thought but I don't really get why sin(a)/a -> 1 as a -> 0.

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(Original post by

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

**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**)
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#10

(Original post by

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

**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

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

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#11

**e-unit**)

Do you need to use Taylor's expansions? If not could someone show me how to do it?

Thanks

dy/dx ≈ lim (f(x+h)-f(x))/(h+x-x) as h-->0

So for sinx

dy/dx = lim

_{h-->0}(sin(x+h)-sinx)/h

dy/dx = lim

_{h-->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

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#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.

Using L'Hopitals rule would create a circular argument.

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#13

(Original post by

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.

**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.

=> lim(sinx/x) = lim(cosx/1)

Then when x -> 0, cosx -> cos0 = 1

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#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!

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#16

(Original post by

And how exactly do you know that (sinx)' = cosx?

**dvs**)And how exactly do you know that (sinx)' = cosx?

I answered for the question lim(sinx/x) not differentiate (original question)

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#18

(Original post by

And how exactly do you know that (sinx)' = cosx?

**dvs**)And how exactly do you know that (sinx)' = cosx?

*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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