which the base line OX is drawn, this is the same as saying that the area of the polygon is independent of this distance, which evidently must be true. In fact, if we suppose the base line OX to move up or down, remaining parallel to itself, it will add and subtract equal areas to or from the positive and negative trapezoids, and so will leave the algebraic sum of the areas unchanged.

Now let us suppose p zero, and put

y1 = a sin a;

y1 = 2a sin a+b sin ẞy,+ a sin a+b sin ß;

Y2

=

Y1 = 2a sin a + 26 sin ẞ+c sin y=y, +b sin ẞ+c sin y; (b) y.y,+c sin y+d sin &;

=

Yo = y.+d sin de sin ε.

(We remark that y, will come out equal to — e sin & if everything is correct.)

We shall then have, for the double of the area of the polygon, 2 Area ya cos a+y,b cos ẞ+y,c cos y+y,d cos d+ye cos ε. (4)

=

As an example, let us compute the area of the polygon investigated in the example of § 72. The following table shows the principal parts of the computation:

The first column gives the values of a sin a, b sin ẞ, c sin y, etc., already computed in the preceding example.

The second column gives the values of a sin a + b sin ß, b sin 6c sin y, etc., which are added to each value of y to form the value of y next following, as shown in (b).

The third column gives the values of y1, Y., Y., Y., Y., computed by the formulæ (6), from the numbers in the first two columns.

The fourth column gives the values of a cos a, b cos ß, c cos y, etc., already computed.

The fifth column gives the products which enter into the equation (c).

The algebraic sum being twice the area, we have

Area of polygon = 1644.1 square metres.

74. It will be noticed that the area of the polygon comes out negative, and the question arises, What interpretation is to be put on this result? The answer is that the conventions of positive and negative as employed in this chapter, as applied to lines, express only the directions in which the lines are reckoned. By a change of direction the algebraic signs of all the trapezoids, and therefore of the area of the polygon itself, will be changed. A little consideration will show that this area will come out positive when we go round the polygon in what we have called the negative direction, and vice versa.

The algebraic sum of the areas, whether positive or negative, will always be the true area of the trapezoid under one important condition: that none of the sides cross each other. In this case the system of applying the algebraic signs will lead to the areas on the two sides of the point of crossing having opposite signs. Hence the result finally obtained will be the difference of the two areas.

EXERCISES.

1. If the lengths and directions of three of the four sides of a quadrilateral are

a 262.72 metres; α = 39° 49';

b = 109.79 66 B = 150° 26';

2. The sides of a quadrilateral, taken in regular order, have the following lengths and directions, in part:

3. Prove geometrically that if the sides of a polygon be joined together in any order, their directions remaining unaltered, the end of the last side will still fall upon the beginning of the first.

An example of the construction is shown in the figure, where the sides are changed from the order abcd to adbe.

4. Express the lengths (x and x')

and the directions ( and ') of the diagonals of a quadrilateral of which the lengths of the sides, taken in order, are a, b, c, d, and their directions a, ẞ, y, d.

δ.

It is only necessary to express the values of the quantities x sin 0, x cos 0, a' sin ', and ' cos ' in terms of a, b, c, d, a, ß, etc. We may suppose a and to refer to the diagonal from the beginning of a to the beginning of c, and ♫ and

to run from the end of a to the end of c.

5. Using the same notation as in § 71, prove

b sin (6a)+c sin (y

a+b cos (ẞ — a) + c cos (y

a) + d sin (♂ — a) + etc. = 0;

a) +'d cos (8 — a) + etc. =

0.

6. If a, ß, and y are the angles which the sides a, b, and c of a triangle make with a base line, and A, B, and C are the interior angles of the triangle, it is required—

(1) To show

A = 180° + B − y ;

B=180° + y — α ;

C=180° + a — - ß. (2) By combining the equations of Ex. 5, to deduce the law of sines (§ 56, 3) and the fundamental equations (§ 58, 9).

7. From the same point O emanate three lines, OA, OB, OC, of such lengths and directions that the sum of their projections upon any third line vanishes. If we complete the three parallelograms, of each of which two adjacent sides are two of the lines, the areas of these three parallelograms will be equal.

Both a geometric and an algebraic proof may be given; the former from § 70, Th. I., the latter from § 62.

8. From the corners A and B of a pentagonal field ABCDE an engineer measures angles as follows: Angle BAC= 79° 23'.6;

47° 29'.7;

Angle ABC =

AB measures 192 metres. Find the remaining sides and the area.

CHAPTER VII.

TRIGONOMETRIC DEVELOPMENTS.

75. LEMMA.

When an arc becomes indefinitely small, the ratio of the sine to its arc approaches unity as its limit.

Remark. In this lemma it is supposed that the arc is expressed in terms of radius as unity.

Proof. It is laid down as an axiom of elementary geometry

that when the number of sides of an inscribed regular polygon is indefinitely increased, the perimeter of the polygon approaches the circumference of the circle as its limit. That is, the ratio

perimeter of polygon

circumference of circle approaches 1 indefinitely.

Now, if we divide the circumference into n equal arcs, the n chords of these arcs, forming the perimeter of the inscribed polygon, will each be twice the sine of half the arc (§ 18). That is, we shall have

Perimeter of polygon = 2n X sine of each arc;
Circumference of circle = 2n equal arcs.

Therefore the above ratio is equal to

sine of arc

arc itself'

* The study of this chapter requires a knowledge of so much of series and im aginary quantities as is contained in Book XI., Chapters I., II., and V., and Book XII., Chapter I., of the author's "College Algebra." It may be advantageously read in connection with Book XII., Chapter II., of that work. Students taking a partial course may pass to spherical trigonometry without reading this chapter.

which therefore approaches unity as its limit when n is increased

indefinitely.

76. PROBLEM. To develop the sine and cosine of an angle in terms of the ascending powers of the angle.

Let us suppose

(1)

(2)

sin x 8,8,x + 8,x2 + 8,x3 +8,x* +8,π° + etc.; cos x = c2+ c1x+€2x2 + ¢ ̧ï3 + c1x2 + c ̧x3 + etc. in which 80, 81, 82, etc., and co, C1, C2, etc., are coefficients whose values are to be determined from some known properties of the sine and cosine.

=

Since sine (-x) sinx, the series for sin x must change its algebraic sign when the algebraic sign of x changes. But only the odd powers of x will then change their sign. Therefore the even powers must not enter into the development, and we must have 8 = 82 = 8, etc. 0.

0

The complete analytic proof of this proposition may be put into the following form. Changing a to x in the development (1), we have sin(x)=80 — 81X + 82X2 — 83x3 + 84x4 — etc.

-

--

But we have, by changing the signs of both members of (1),

Because these developments must be identically equal, we must have

which gives 3。 = 89 = 84 = etc. = 0.

Therefore the development of the sine becomes

sins, 8,1⁄23 +8,x2 + etc.

Dividing this equation by x, we have

Now suppose to approach indefinitely near to zero. The first member will then, according to the lemma, approach unity as its limit, and the second member will approach 8, as its limit. Therefore we must have 8, 1, and the development becomes

=

sin x = x+8,1⁄23 +8x3 + etc.

(3)

Next take the development (2) for cosa. If we suppose x = 0,