695. To express any vector-quotient as a complex number.

Let (OH) and (OA) be any two vectors perpendicular to the axis Z.

From H draw HB perpendicular to OA.

Then (OH) = (OB) + (BH).

Then, since (OB) is in the same line as (OA);

Here a and b are real: i.e. a is positive or negative according as (OB) is along or opposite to (04): and b is positive or negative according as (BH) makes with (OA) an angle + 1⁄2π or - 1⁄2π.

construction for determining a and b is Hence any equation such as

Now the above

absolutely unique.

requires that a

696. Moreover, if = angle between (OA) and (OH),

==

697. To find the cosine and sine of the sum of two angles.

Since cosi, sin and cos +i, sino denote respectively the operations of turning a vector through angles 0 and & in the plane whose axis is Z,

.. cos (0 + $) + i1⁄2 sin (0 + $) = (cos 0 + i2 sin 0) (cos $ + ¿1⁄2 sin $). But, by Art. 690, we may use the distributive law in the multiplication of these versors. Hence

(cos + i sin 0) (cos + i sin o)

= cos 0 cos + iz sin 0 cos

+ i2 cos 0 sin - sin 0 sin ø,

for iii.e. Z′′/2 × Z′′/2 = Z′′ 1. Hence

cos (0 + $) + i sin (0 + $)

and

==

= cos cos - sin 0 sin & + ¿1⁄2 (sin @ cos & + cos @ sin 4). Hence, by Art. 695,

cos (0 + $) = cos cos - sin 0 sin & :

sin (0+) = sin @ cos & + cos ◊ sin &.

This proof is in principle the same as that given in Art. 343. 698. If in the above we put o - 0, we have

(cos + i sin 0) {cos (−.0) + i2 sin (− 0)} = cos (0 – 0) + i1⁄2 sin (0 − 6). But cos - 0) = cos 0; sin (— 0)

-=

- sin 0; cos 0 = 1; sin 0 = 0.

.. cos2 + sin2 0 = 1.

This is a proof of Euc. I. 47. For the reader will perceive that this proposition has not been surreptitiously assumed anywhere.

699. From the above result De Moivre's theorem follows at

once.

For cos i sin denotes the operation of turning through an angle 0.

.. (cos 0 + i2 sin 0)” denotes the operation of turning through an angle 0, n times: i.e. through an angle nė.

Ө

.. (cos + i sin 0)" = cos n✪ + i sin no.

It should be pointed out, however, that this equation of operations has no value for the purposes of expanding cos no and sin no, unless we take into account the distributive law proved in Art. 690, by means of which we may 'multiply out' the power on the left-hand according to ordinary algebraic rules. The fundamental proposition upon which the whole of analytical trigonometry rests is the 'addition-theorem' of Art. 697, from which the Demoivrean expansions algebraically follow.

(1) Any vector-quotient is represented by a complex number. (2) Its tensor is the modulus of that complex number.

(3) Its versor is the Demoivrean function of the angle between the two vectors.

701. Since p. Zxq. Z=pq. Z°+, we have the fundamentally important theorem that

The modulus of the product of two complex numbers is equal to the product of their moduli.

702. To expand the cosine and sine of an angle in terms of its circular measure.

Let (OA) be any initial vector.

Let a moving point trace out the path (AB) + (BC) + (CD) + ..., in which the angles OAB, OBC, OCD,... are all right-angles, and

Thus, if M, N are the nth and (n - 1)th points, respectively, of the series A, B, C..., then

ON = p2. OA and MN= kp"-1. OA.

Thus the length of the path traced by the moving point from A to N is

k (1 + p + p2 + ... + p”−1). OA = k (p” − 1). Od.

p-1

Now, let k = 0/n: and let n increase indefinitely, while remains finite, so that k diminishes indefinitely.

Then it will be seen that limit of p2 = 1; and limit of

Hence the tracing point describes a circle of radius OA, and

Now the operation to be performed upon each of the vectors (OA), (OB), (OC)... to obtain the next is 1 + ki.

Increasing n indefinitely and expanding by the binomial theorem, we have

But is the circular measure of the angle AON by (1), and ON=OA:

.. cos + i sin = expansional value of ei.

703. In the above, we have used i for the operator which turns a vector through a right-angle in any fixed plane.

Let Z denote the operation of turning an angle through some undetermined unit of angular measurement in the plane whose axis is Z. And let A be the measure of in terms of this unit. Then

ᎾᎿ

Z1 = cos A + i, sin A = cos 0 + i2 sin 0 = eoi..

The most convenient unit to take will be the right-angle. Then A right-angles = πA radians: or 0 = 1⁄2πA.

704. If X, Y, Z be three axes at right-angles to one another, the operations of turning through a right-angle round these axes will be respectively ix, i, iz. For these three operations

Hamilton used the symbols i, j, k.

705. If we have a fixed plane for rotation, and a fixed direction in that plane for our initial vector, then the operation i upon this initial vector will give a vector fixed in the plane of rotation at right-angles to the initial vector.

where a and b are abstract ratios, having therefore neither length nor direction, we have

(OH) = a. (OA) + bi (OA).

Now, if a. (OA) and b.(OA) be represented by a and ß, respectively, a and ẞ will stand for lengths having an arbitrarily fixed direction. Thus, since

.. any vector in the fixed plane is expressed in terms of its projection and erection upon any fixed line in that plane.