Michael Ede

Posted on Monday, 19 January, 2004 - 04:08 pm:

I need to differentiate sin[x] from first principles. I have got the beginning but cannot prove the end.

y=sin[x]

y+dy = sin(x+dx)
y+dy = sin[x]cos[dx]+cos[x]sin[dx]
dy = sin[x](cos[dx]-1)+cos[x]sin[dx]

dy/dx = (sin[x](cos[dx]-1))/dx+(cos[x]sin[dx])/dx

Now (cos[dx]-1)/dx must tend toward 0 and (sin[dx])/dx must tend toward 1, but I cannot prove this.

Ian Short

Posted on Monday, 19 January, 2004 - 04:24 pm:

Do you know the power series expansions, the Maclaurin series?

sinx =x-x³ /3!+x⁵ /5!+...

and one for cosine by differentiating the above thing term by term. These will help.

Ian

Demetres Christofides

Posted on Monday, 19 January, 2004 - 05:03 pm:

Alternatively, consider the unit circle with centre O = (0,0). Let A = (1,0) and consider the line segment OB where B is a point on the circle such that the angle BOA = x (in radians). [We'll consider only 0< x < pi/2]. By definition, the arc AB has length x. Bring the perpendicular from B to OA meeting OA at D. Then

x = arc AB > = BA > = BD = (OB)sinx = sinx

So (sinx)/x < = 1

No draw a parallel CA to BD meeting OB at C.

Then sinx = AC/OC > = x/OC = x/(OB + BC) = x/(1+BC)

So 1/(1+BC) < = sinx/x < = 1

As x becomes very small, BC becomes very small too and so sinx/x tends to 1.

Demetres

Kerwin Hui

Posted on Monday, 19 January, 2004 - 06:07 pm:

Also, you can avoid computing the limit of

(\cos (δ x) - 1) / δ x

by using the sum-to-product formula instead of the compound angle formula for sine. That is,

$\sin (x + δ x) - \sin x = 2 \cos (x + δ x / 2) \sin (δ x / 2)$

So $[\sin (x + δ x) - \sin x] δ x = \cos (x + δ x / 2) [\sin (δ x / 2) / δ x / 2)]$ .

Kerwin

Michael Ede

Posted on Tuesday, 20 January, 2004 - 07:00 pm:

Thanks, I understand the sin bit, but I still can't show that (cos(dx)-1)/dx tends to 0.

Kerwin Hui

Posted on Tuesday, 20 January, 2004 - 08:18 pm:

You don't have to. See my post above.

If you are differentiating cos(x), then using the sum-to-product formula (together with the limit sin(h)/h tends to 1 as h tends to 0) will give you d(cos x)/dx=-sin(x), which implies that (cos(x)-1)/x tends to 0 as x tends to 0.

Kerwin

Michael Ede

Posted on Tuesday, 20 January, 2004 - 08:28 pm:

Ah, I understand. Thanks a lot.