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NOTE. The representation of the trigonometric functions by lines is for the sake of clearness. They are not really lines, but ratios of lines which are pure numbers. But in studying these numbers the ideas are fixed by representing them by lines, as is done in some departments of algebra. We have only to remember that the lines are not the functions themselves, but lengths proportional to the functions, and therefore admitting of being used to represent the functions. These lengths are, however, really equal to the products of the radius by the corresponding functions. For example, if

sin XOM =

PM
Radius'

then

sin XOM. Radius = PM.

18. Remark. The sine of an angle is equal to half the chord of twice the arc of the angle, the radius being supposed unity. Hence :

Any chord in a circle is equal to the radius multiplied by twice the sine of half the angle subtended by the chord.

EXERCISE.

Let the student find by actual measurement with dividers and scale the sine, tangent, and secant of every 10° from 0° to 90° in the following way:

With a radius equal to some unit or some whole number of units on a scale, describe the quadrant XB. Either 4 inches, 5 inches, or a decimetre would be a convenient radius.

Divide the quadrant into 9 ares of 10° each. Through each point of division, M for instance, draw a radius and continue it until it intersects the tangent at N. Then

measure

B

X

N

1. The distance of each division-point on the arc from the line OX, which distance, divided by the radius, will give the sine of the corresponding angle.

2. The distance of each point of intersection, N, from X, which being divided by the radius will give the tangent of the angle.

3. The length of each ON, which being divided by the radius will give the secant of the corresponding angle.

The results should all be expressed in decimals to three places, exhibited in a little table in the following form, and afterward compared with the values found in the trigonometric tables :

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With care the average deviation of the measures from the truth ought not to exceed .005, except in the cases of the tangent and secant of 70° and 80°, which are so great that they cannot be easily found in this way.

19. To find, by measurement, the angle corresponding to a given sine, tangent, or secant.

Analysis. If a sine is given, the end of the arc corresponding to the required angle must be at a distance from the line OX equal to the given sine, the radius being unity. y Therefore if we take on the perpendicular OY a distance OR equal to the product of the radius by the given sine, and through R draw a parallel to OX, the point in which it intersects the arc will give the required angle.

To find the angle corresponding to a given

R

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tangent we take a distance XS equal to the product of the given tangent into the radius. Join OS. The angle XOS will be that required.

For a secant we take the product of the radius by the secant in the dividers, and from O as a centre draw an arc cutting XN in a point N. Joining ON, the angle XON will be that required.

EXERCISES.

1. Find by measurement the angles of which the sines are 1, 0.3, 0.4, 0.6.

2. Find arc-tan, 1, 1.5, 3.

The expression arc-tan is used for brevity to mean the arc corresponding to a given tangent.

3. Find arc-secant 1.5, 2.

20. Functions of unlimited angles. Thus far we have con sidered only the sines, tangents, and secants of angles less than 90°; that is, of angles in the first quadrant (§9). As our angle increases to an entire cir- M cumference, the functions are deter- U mined by the same construction modified to suit the case.

The following are the general definitions:

We first generalize the construction.

M

On the sides of the angle we take equal lengths OX and OM

as the unit of measurement and radius of the circle.

At

we erect a line TXT' perpendicular to OX, extending indefinitely in both directions.

We also suppose the revolving side OM to be produced indefinitely in both directions, and XO to be produced so as to form the diameter XU.

Then, however the side OM may revolve

I. The sine of the angle XOM is always represented by the perpendicular from M upon the line OX. The sine is positive or negative according as M is above or below OX.

II. The tangent of the angle XOM is always represented by the distance from X to the point in which the side OM produced intersects the line TXT'. The tangent is positive or negative according as the point of intersection is above or below X.

III. The secant of XOM is the length of OM produced intercepted between 0 and the vertical line TXT'. The secant is

positive or negative according as it is in the direction from O toward M or in the opposite direction. The positive direction of OM is supposed to revolve with the side OM, and therefore to be always from O toward M.

Y

21. Changes in the value of the sine. If we suppose the side Om of the angle XOm to revolve around 0, carrying the sine mp with it, the latter will increase to its maximum value, equal to radius unity, when m reaches Y, and angle XOm = 90°. Hence

sin 90° +1.

As m moves from Y to X', the sine will diminish from 1 to zero. Because

angle XOX' = 180°,

sin 180° = 0.

M

m

P'

X

P

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X

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If m passes over X' into the third quadrant, the perpendicular M'P' will be below the line X'OX. This change of direction is expressed by changing the algebraic sign of the perpendicular from

to. This is in accordance with the following general principle: Whenever distances measured in one direction are considered positive, those in the opposite direction are negative.

Hence also:

In the third quadrant the sine is negative.

When the point m reaches the position Y' it will have moved through three quadrants or 270°, and the sine will coincide with the radius OY' of length unity. Hence

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As m moves from Y' to X, the sine will increase from to 0. Hence:

tive.

I. In the fourth quadrant the sine is nega

II.

sin 360° sin 0° = 0.

The changes of algebraic sign as the angle goes through the four quadrants are shown in the annexed diagram.

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Angles having equal sines. If angle X'OM = XOm, the

two angles XOm and XOM will be supplementary. Also in this case the triangles XOm and X'OM will be identically equal; so that

PM=pm.

Now PM represents by construction the sine of XOM, and pm the sine of XOm. Hence:

The sines of supplementary angles are equal.

In symbolic language this theorem is expressed thus: If a be any angle, then

and

a) = sin a

sin (180° — a):

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(2)

If the points M' and M" are equally distant from Y', so that angle M'OY' = angle Y'OM", which angle call y, the sines P'M' and P''M" will be equal. Hence, whatever be y,

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22. Changes in the tangent. As the line Om revolves round O and m approaches Y, the point of intersection N will move upwards without limit. As m reaches Y, Om will become parallel to the tangent line, and the point N will recede to M infinity. Hence :

The tangent of 90° is infinite.

X

Y

m

X

When m is in the second quadrant, suppose in the position M, the revolving side OM will not intersect the tangent line at all in the positive direction OM. We must therefore suppose the revolving line to be produced in the negative direction ON' so as to intersect the tangent line at N' below X. The distance XN' is then to be regarded as negative. Hence:

In the second quadrant the tangent is negative.

Following the motion, we see that when m reaches X', reaches and the tangent becomes zero. Hence

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N'

When m is in the third quadrant, N passes above X and the tangent is positive, so that

In the third quadrant the tangent is positive.

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