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Trigonometric Power Series
By Zhidong Leong on Wednesday, July 18, 2001 - 05:42 pm:

What are the power series expansions for trigonometric functions such as sin, cos and tan?

By Brad Rodgers on Thursday, July 19, 2001 - 04:19 pm:

I'm assuming that this is what you mean:

sin(x)=x-(x3/3!)+(x5/5!)-(x7/7!)+...

cos(x)=1-(x2/2!)+(x4/4!)-(x6/6!)+...

tan(x)=sin(x)/cos(x)

There is also a series for tan(x), but it is uneccesarily complicated. These series give accurate values when x is measured in radians. I can prove these formulas, but it takes calculus. If you want the proof, though, just ask.

A couple of far less useful formulas are

cos(x)=(eix+e-ix)/2

and

sin(x)=(eix-e-ix)/2i

If that's not what you wanted, don't hesitate to post again.

Brad

By Zhidong Leong on Friday, July 20, 2001 - 11:22 am:

Yes, that's what I wanted. Thanks a lot.
But why is it in series form as
its graph is a smooth and nice curve.
Why isn't it in quadratic form?

Can I have the proofs?

[Editor: Not only polynomials have 'smooth' looking graphs. Arun explains how the series expansion arises below. A complete understanding of the situation comes through appreciating Taylor's Theorem, which is typically taught in a first course in abstract analysis (first year British maths undergraduate course).]
By Arun Iyer on Friday, July 20, 2001 - 08:18 pm:

The proofs you require are as follows...
The parts you need to know about are derivatives of functions. The derivative of f(x)
evaluated at x = x0 gives the slope of the tangent line to the curve
y = f(x) at the point (x0,f[x0]). The key facts are these:

1. The derivative of sin(x) is cos(x).
2. The derivative of cos(x) is -sin(x).
3. The derivative of xn is nxn-1 for any integer n.
4. The derivative of cf(x) is c times the derivative of f(x), for any constant c.
5. The derivative of f(x)+g(x) is the sum of the derivatives of f(x) and g(x).

Now assume that sin(x) (for example) has a convergent power series expansion, that is,

sin(x) = a(0) + a(1)x + a(2)x2 + ... + a(n)xn + ...

Substitute x = 0 in this equation, and you get a(0) = 0. Now take the derivative of both sides:

cos(x) = a(1) + 2a(2)x + 3a(3)x2 + ... + na(n)xn-1 + ...

Substitute x = 0 in this new equation, and you get a(1) = 1. Again
take the derivative of both sides:

-sin(x) = 2!/0!a(2) + 3!/1!a(3)x + 4!/2!a(4)x2 + ...

Substitute x = 0 in this new equation, and you get a(2) = 0. Again take the derivative of both sides:

-cos(x) = 3!/0!a(3) + 4!/1!a(4)x + 5!/2!a(5)x2 + ...

Substitute x = 0 in this new equation, and you get a(3) = -1/3!. Again take the derivative of both sides:

sin(x) = 4!/0!a(4) + 5!/1!a(5)x + 6!/2!a(6)x2 + ...

Substitute x = 0 in this new equation, and you get a(4) = 0.

If you repeat this, you will see (and can prove by induction) that the coefficients are given by

a(2k) = 0
a(2k+1) = (-1)k/(2k+1)!

for every integer k. That means that the series has form

sin(x) = x - x3/3! + x5/5! - x7/7! + x9/9! - ...

By taking the derivative of this, you get the series for cosine:

cos(x) = 1 - x2/2! + x4/4! - x6/6! + x8/8! - ...

More details of this development you will encounter when you study infinite power series in calculus. It is studied under the title, 'Maclaurin's Series'.

If you need more help, feel free to write back.

love arun