This page deals with the proof of De Moivre's Theorem, etc. It has formula to compute the cosine and sine directly, but these require further algebraic manipulation, to reduce the sines in the cosine formula, and the cosines in the sine formula. This work is extended later beginning with the cosine, where formulae are also derived to calculate directly the coefficients of a given power in a given expansion for cos nx and sin nx. Before that, some pages, mentioned below, deal with the Chebyshev method.

See also

- Chebyshev's Method
- Multiple Angles Cosines
- Multiple Angles Sines
- Multiple Angles Tangents
- De Moivre's Theorem Extended

Trigonometry Contents

[1.1]

where i=√(-1)

Picking out the real parts from both sides and equating them, for cosine, we get:

[1.2]

Where k, a nonnegative integer, is 2·p, for the pth term, p=0, 1, 2..., and k≤n

For the sine, we find:

[1.3]

Where k, a nonnegative integer, is 1+2·p, for the pth term, p=0, 1, 2..., and k≤n

We can use this formula to find multiple angles of cosine and sine by associating the real parts of the right-hand side, expanded with the Binomial Theorem, with cosine and the imaginary parts with the sine. We give examples on a following page.

On this page, we make a limited claim, that the formula is true for all integers n. Actually it is true in a much wider context, for complex numbers.

[2.1]

Because

[2.2]

We can substitute ix for x (where i=√-1 )

[2.3]

Replacing i

[2.4]

Grouping the real and imaginary parts:

[2.5]

We note the first series is the expansion of cos x, and the second i·sin x, so:

(which is what Euler claimed)

By raising e

Which, assuming the expansions above for e

■

[1.1, repeated]

Where n is a natural number

When n=1, then:

cos x+i·sin x=(cos x+i·sin x)

Which is true, so the theorem is true for n=1.

For n=k, where k is a nonnegative integer, by the formula, we have

cos kx+i·sin kx=(cos x+i·sin x)

Assuming that it is true for n=k, then when n=k+1:

cos (k+1)x+i·sin (k+1)x=

(cos kx+i·sin kx)(cos x+i·sin x) [3.2]

by multiplying Equation 3.1 by (cos x+i·sin x)

Multiplying out Equation 3.2 we find:

=cos kx·cos x+i

=cos kx·cos x-sin kx·sin x+i·sin kx·cos x+i·cos kx·sin x [because i

=cos (k+1)x+i·sin (k+1)

by the compound angle formula.

So if the formula is true for n=k, it is also true for n=k+1. As it true for n=1, then it is true for all n

■

So:

[4.2]

Applying De Moivre's formula to the right-hand side:

[4.3]

Multiplying top and bottom by cos (mx)-i·sin(mx), noting (a−b)(a+b)=a

[4.4]

And because sin

[4.5]

Substituting n back into the formula, that is, substituting −n for m:

[4.6]

Taking the minuses outside, and noting that cos(−nx)=cos(nx), and sin (−nx)=−sin(nx), we have:

[4.7]

which is De Moivre's formula for negative n, so the formula works for all integers . ■

Trigonometry Contents

Ken Ward's Mathematics Pages

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