# De Moivre's theorem.

• Mar 2nd 2009, 07:27 AM
Erghhh
De Moivre's theorem.
Find tan5x in terms of tanx.

Not sure if I'm meant to manipulate z=cosx+isinx to get z/cosx=1+itanx?
• Mar 2nd 2009, 06:17 PM
Soroban
Hello, Erghhh!

It can be done, but I don't know if my method is what they want.

Quote:

Use DeMoivre's Theorem to find $\tan5x$ in terms of $\tan x$
DeMoivre's Theorem says: . $\left(\cos\theta + i\sin\theta\right)^5 \:=\:\cos5\theta + i\sin5\theta$

So we have: . $\cos5x + i\sin5x \:=\:(\cos x + i\sin x)^5$

. . $\cos5x + i\sin5x \:=\:\cos^5\!x + 5i\cos^4\!x\sin x - 10\cos^3\!x\sin^2\!x -$ $10i\cos^2\!x\sin^3\!x + 5\cos x\sin^4\!x + i\sin^5\!x$

. . $\cos5x + i\sin 5x \:=\:\left(\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x\right) +$ $i\left(5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x\right)$

Equate real and imaginary components:

. . $\cos5x \:=\:\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x$

. . $\sin5x \:=\:5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x$

We have: . $\tan5x \:=\:\frac{\sin5x}{\cos5x} \;=\;\frac{5\cos^4\!x\sin x - 10\cos^2\!x\sin^3\!x + \sin^5\!x} {\cos^5\!x - 10\cos^3\!x\sin^2\!x + 5\cos x\sin^4\!x}$

Divide top and bottom by $\cos^5\!x\!:\;\;\tan 5x \;=\; \frac{5\,\dfrac{\sin x}{\cos x} - 10\,\dfrac{\sin^3\!x}{\cos^3\!x} + \dfrac{\sin^5\!x}{\cos^5\!x}} {1 - 10\,\dfrac{\sin^2\!x}{\cos^2\!x} + 5\,\dfrac{\sin^4\!x}{\cos^4\!x}}$

. . Therefore: . $\boxed{\tan5x \;=\;\frac{5\tan x - 10\tan^3\!x + \tan^5\!x}{1 - 10\tan^2\!x + 5\cos^4\!x}}$

• Mar 2nd 2009, 07:23 PM
Soroban

We never have to go through all that ever again!
These multiple-angle formulas can be written with a rhythmic pattern.

Suppose we want: . $\sin6x,\;\cos6x,\;\tan6x$

Expand: . $(\cos x + \sin x)^6$
. . $\cos^6\!x + 6\cos^5\!x\sin x + 15\cos^4\!x\sin^2\!x + 20\cos^3\!x\sin^3\!x + 15\cos^2\!x\sin^4\!x +$ $6\cos x\sin^5\!x + \sin^6\!x$

Write these terms alternately in the denominator and numerator of a fraction.

Start with the denominator: . $\frac{\qquad6\cos^5\!x\sin x \qquad 20\cos^3\!x\sin^3\!x \qquad 6\cos x\sin^5\!x}{\cos^6\!x \qquad 15\cos^4\!x\sin^2\!x \qquad 15\cos^2\!x\sin^4\!x \qquad \sin^6\!x}$

Insert alternating signs in the numerator and in the denominator:

. . . . . . . $\frac{6\cos^5\!x\sin x \:{\color{red}-}\: 20\cos^3\!x\sin^3\!x \:{\color{red}+} \:6\cos x\sin^5\!x}{\cos^6\!x \:{\color{red}-}\: 15\cos^4\!x\sin^2\!x \:{\color{red}+}\: 15\cos^2\!x\sin^4\!x \:{\color{red}-}\:\sin^6\!x}$

And we have two of the multiple-angle identities:

Numerator: . . $\boxed{\sin 6x \;=\;6\cos^5\!\sin x - 20\cos^3\!x\sin^3\!x + 6\cos x\sin^5\!x}$

Denominator: . $\boxed{\cos6x \;=\;\cos^6\!x - 15\cos^4\!x\sin^2\!x + 15\cos^2\!x\sin^4\!x - \sin^6\!x}$

For $\tan6x$ divide top and bottom of the fraction by $\cos^6\!x$

. . $\boxed{\tan6x \;=\;\frac{6\tan x - 20\tan^3\!x + 6\tan^5\!x}{1 - 16\tan^2\!x + 15\tan^4\!x - \tan^6\!x}}$