when four straight lines are proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means (VI. 16); consequently, the arithmetical product of the extremes is equal to that of the means, or, which is the same thing, the quotient of the first antecedent by its consequent is equal to that of the second antecedent by its consequent.

Retaining, therefore, the hypothesis that the radius is 1, we obtain from the proportions given in Art. 10,

1

(3); cot A=;

cos A

1

tan A

(4);

cosec A=- (5); sin 2A+cos 2A=1 (6);

sin A

1+tan 2A=sec 2A (7); 1+cot 2A=cosec 2A (8).

14. PROP. VII.

To exhibit the sine and cosine arithmetically.

By similar triangles (Fig. Prop. V.), CD: CB:: NO : BN;

CD

NO

Therefore (arithmetically) CBBN

BN
BO

=

=

hypotenuse

base

Similarly cos A=BN hypotenuse

These expressions are employed by many writers as the defimtions of the sine and cosine; and the circle is altogether dispensed with.

15. EXAMPLE 1.—To find the arithmetical values of the sine, cosine, and tangent of 45o.

Let ABC be an isosceles right-angled triangle, BC=1. Then BCA-45°, and sin BCA-AB. Now AB2+AC2=BC2 (I. 47) .=1; or 2AB2=1; therefore, sin 45°=AB=

1

B

A

16. EXAMPLE 2.—To find the arithmetical values of the sine, cosine, and tangent of 60° and 30.°

C

Let ABC be an equilateral triangle; AB=1. Draw CD perpendicular to the base AB; then it bisects AB (1. 26). Hence cos 60°=cos ABC=BD=1BA=1, and BD+CD2-CB2=1 (I. 47); which gives +

No3

CD2 = 1, and CD2 = 2, or CD=- ; therefore

2

Now, sin 30°=cos 60°=}; cos 30°=sin 60°=3; and, there

2

E

17. EXAMPLE 3.—To find the sine of 18°, or cosine of 72°. The triangle which is constructed in Prop. 10, Book IV. is such that B=D=2A; and, consequently, all the angles together are equal to 5A; therefore, 5A=180° (I. 32); or A=36°. Let AB=1; then, by construction, AC2=AB.BC; whence AC2= BC, and AC2+AC=BC+AC=AB=1. √5-1 Now when squared and added to itself 2

gives unity; therefore, AC="

√5-1; and sin 18°

2

B

√5-1

4

or cos 72° chord 36° (Art. 6) =1BD=AC="

† 18* PROP. VIII.

A

To ascertain the changes of sign of the trigonometrical functions. For the purpose of generalization, that is, of rendering expressions which have been obtained from one particular figure applicable to every case which the enunciation is capable of including, it is necessary that some of the trigonometrical functions should, for certain values of the angle, be negative in sign. To ascertain when this will happen, relative to the sine and the cosine, we must examine how the values of these functions vary as the angle varies. We perceive, then, that as the angle increases from 0 to 90°, the sine also increases; and as the angle increases from 90° to 180, the sine diminishes. But, in every case, the line which represents the sine is drawn in the same direction from the revolving radius to the fixed one. When the angle is greater than 180°, it is not so; this line is drawn in the opposite direction, and must, therefore, be negative in sign.

* The student should omit all articles marked † at the first reading.

Again, the cosine diminishes as the angle increases from 0° to 90° in such a manner that cos 0°=1, and cos 90°-0. As the angle increases from 90° to 180°, the cosine is drawn in the opposite direction, and must be negative in sign. It may be remarked that, when the figure is drawn as we have drawn it (Art. 4), the circle is divided into two semicircles by a horizontal line, in the upper of which the sine is positive, and in the lower negative; but it is divided into two semicircles by a vertical line, in the further of which the cosine is positive, and in the nearer, negative.

Having thus determined the sign of the sine and the cosine, the relations exhibited in Art. 10, enable us to determine the signs of the other trigonometrical functions. For, since tan A: R:: sin A: cos A, the tangent will be positive when the sine and the cosine have the same sign, but negative when they have contrary signs. The same is true of the cotangent. Also, since sec A: R::R: cos A, the secant will have the same sign as the cosine. In like manner it may be shown that the cosecant will have the same sign as the sine. Lastly, the versed sine is manifestly always positive. These signs are exhibited in the following table :

† 19. COR. It has been shown in Art. 9, that (abstracting from sign) cos (90+A)=sin A &c. We are now enabled to write the values of these expressions with their proper sign; thus, cos (90+A) = sin A, sin (90+A)= cos A; sin (180-A) = sin A; cos (180-A) cos A; sin (180+A) = −sin A; cos (180+A)= -cos A.

=

==

SECTION II.

PROPERTIES OF TWO OR MORE ANGLES.

20. PROP. I.

Given the sines and cosines of two angles, to find the sines and cosines of their sum and difference.

Let ABC-A, CBD=B, be the two angles, of which A is the greater.

Then, in Fig. 1, ABD is the sum of A and B; and in Fig. 2, ABD is their difference. With radius BA=R, describe the circle ACD. Draw DE, DF at right angles to BA, BC; FG, FH at right angles to BA, DE, and CK at right angles to BA.

Then (Fig. 1) rectangle R sin (A+B) = BC.DE=BC.EH + BC.HD (II. 1): BC.GF+BC.HD

= B

And by similar triangles, BC: KC::BF: GF, therefore (VI. 16) BC.GF=KC.BF. Moreover, the angles ABC, HDF are equal, being both the complements of the angle between BC and ED; hence the triangles BGF, DHF are similar; therefore BC: BK:: FD: HD; and, consequently (VI. 16), BC.HD=BK.FD; B whence

R sin (A+B)=KC.BF+BK.FD

sin A cos B+cos A sin B.

H

Fig. 1.

D

E

GK

A

Fig. 2.

F

Again (Fig. 2), R sin (A—B)=BC.DE=BC.EH-BC.HD

-BC.GF-BC.HD

=KC.BF-BK.FD

=sin A cos B-cos A sin B.

In like manner (Fig. 1), R cos (A+B)=BC.BE =BC.BG-BC.EG=BC.BG-BC.HF.

But by similar triangles, BG: BF :: BK: BC; therefore BC.BG =BK.BF; also, HF: FD:: CK: BC; therefore BC.HF=CK.FD; consequently, R cos (A+B)=BK.BF-CK.FD

=cos A cos B-sin A sin B.

Also (Fig. 2), R cos (A-B)=BC.BE=BC.BG+BC.EG

=BC.BG+BC.HF

=BK.BF÷CK.FD

=cos A cos B+sin A sin B

21. The preceding proposition may be proved more simply by adopting the definitions given in Art. 14; for then the circle is altogether dispensed with, and all that is required is to reduce the two fractions, whose sum or difference makes up the sine or cosine, into compound fractions, by introducing in each, as a denominator, the hypotenuse of the triangle to which the numerator belongs. Thus,

22. COR. Given the sine and cosine of A, to find those of 2A.

Let B-A in the last proposition;

then, sin 2A=sin A cos A+cos A sin A,

=2 sin A cos A (1);

cos 2A= =cos A cos A-sin A sin A, =cos 2A-sin 2A (2) ;

or=1-sin 'A-sin 'A (Art. 13), =1-2 sin 2A (3);

or =2 cos 2A-(cos 2A+sin 2A), =2 cos 2A-1 (4) (Art. 13).

23. LEMMA.-If there be two unequal magnitudes, half their difference added to half their sum is equal to the greater; and half their difference taken from half their sum is equal to the

less.

Let AB and BC be two unequal magnitudes, of which AB is the greater; suppose AC bisected in D, and AE equal to BC. It is manifest that AC is the sum, and

E D B

C

EB the difference of the two mag- A nitudes. And because AC is bisected in D, AD is equal to DC; but AE is also equal to BC, therefore DE is equal to DB, and DE or DB is half the difference of the magnitudes. But AB is equal to BD and DA, that is to half the difference added to half the sum; and BC is equal to the excess of DC, half the sum, above BD, half the difference. Q. E. D.

COR. 1. Hence, if from half the sum of two magnitudes, the less be taken, there will remain a magnitude equal to half their difference.

COR. 2. Hence, also, if from the greater of two magnitudes half the sum be taken, there will remain a magnitude equal to half the difference.