34. In the triangles CMP and CED, which have their sides respectively parallel, and are therefore similar, we have the proportion,*

36. In the expressions for the tangent and cotangent which we have here derived, it will be observed that we have the quotient of the sine and cosine, and that therefore when the sine and cosine have contrary signs, the tangent and cotangent will be negative. This occurs in the second and fourth quadrants.

It appears hence, that the cotangent changes its sign always with the tangent.

Also that both the tangent and cotangent of an arc are equal to those of its supplement with contrary signs.

From the expressions for the secant and cosecant, it appears that the former must always have the same sign as the cosine, and the latter the same as the sine.

The formulas derived in the last four articles should be committed to memory.

Quantities in changing their signs pass through zero or infinity. (See Alg. Note 3d, p. 176.) Thus the sine changes from to or vice versâ, twice in going round the circle; viz. in passing through 0 at 0° and 180°; the cosine twice in passing through 0 at 90° and 270°; the tangent four times, in passing through 0 at 0° and 180°, and at 900 and 270°; the cotangent four times, in passing through 0 at 90° and 270°, and through infinity at 0° and 180°; the secant twice, in passing through at 90° and 270°; the cosecant twice, in passing through ∞ at 0° and 180°.

37. Multiplying the expression for the tangent given in Art. 32 by that of the cotangent in Art. 34, we have

i. e. the tangent and cotangent are reciprocals of each other.

EXERCISES.

1. Express each of the six trigonometrical lines in terms of each one of the other five by itself.

The sine of an angle being 0•856, find the other trigonometrical functions.

3. The tangent being 2.34, find the others.

4. The cotangent being 1.203.

5. State the equivalents of the following functions of angles greater than 90°, or obtuse angles in equivalent functions of angles less than 900 or acute angles, sin 170, cos 1470, tan 98° 31', cot 1710 14', sec 1710, cosec 1550, sec 2155, cos 3188 10', tan 271, 81, cot 2048 18` 94.

6. Find the versed sine of the angle, the cosine of which is 93358.

7. Prove the vers. 1 cos to be always positive.

8. Prove the greatest value of the versed sine to be 2R.

38. We are now prepared to find formulas for the solution of right angled plane triangles in all cases, and plane triangles in general in a few particular ones. The remaining cases of triangles in general will require further preliminary matter.

DERIVATION OF FORMULAS FOR THE SOLUTION OF RIGHT ANGLED PLANE

perpendicular to CN. MP will be the sine of the arc MN because it is drawn from one extremity м of the arc perpendicular to the diameter which passes through the other extremity N.

MP is also the sine of the angle c. CM is the radius of the circle to which the arc MN belongs.

The two triangles CMP and CBA are equiangular and similar, and give the proportion

CM: MP :: CB: BA or R: sin c :: CB: BA.

Had an arc been described with в as a centre in a similar manner we

should have had R: Sin B:: BC: CA, from which it appears that the radius of any circle whatever* bears the same proportion to the sine in that circle of the arc which measures one of the acute angles of a right angled triangle, that the hypothenuse of the triangle does to the side opposite the acute angle.

It is customary, for conciseness, to represent the sides opposite the angles of a triangle by small letters of the same name with the large letters which are placed at the angles; which large letters are also employed as the algebraic representatives of the angles. Thus in the triangle above, A being the right angle, the hypothenuse opposite is expressed by a; the side AC opposite в is represented by b, and so the other. The above proportions would, according to this method, be written thus

Both these proportions are expressed in the single rule printed in italics above. When R = 1, multiplying the second and third terms, and dividing by the first, in the preceding proportions we have

and

b = a sin B

c = a sin c

(2)

=

That is either perp. side the hypoth. X the sine of the angle opposite. The two acute angles of a right angled triangle are together equal to a right angle or 90° (Plane Geom. Theorem 15, Cor. 5), therefore they are complements of each other; hence sin c = cos B; and the second of the above proportions (1) may be changed into

which may be translated into ordinary language thus; radius: the cosine of one of the acute angles of a right angled triangle: : the hypothenuse: the side adjacent the acute angle.

When any three terms of a proportion are given, the remaining term can be found. If the unknown term be one of the extremes, multiply

* It is important to observe that the same trigonometrical lines of angles or arcs containing the same number of degrees in two different circles bear the same relation to each other. Thus in the diagram above, CM: MP: CM: MP, or,

also,

(R: sin) of the smaller cire :: (R: sin) of the larger;
CM: CP:: CM: CP or (R: cos) of the one:: (R: cos) of the other.

the two means and divide by the other extreme; if the required term be a mean, multiply the two extremes, and divide by the other mean. When R = 1 we have from proportion (3) above

i. e. either perpendicular side of a right angled triangle, equal to the hypoth. X cos of the adjacent angle.

The above formulas contain each of them two of the sides of a triangle, the sine or cosine of an angle, and radius. If the lengths of the sides be given in numbers, these numbers may be put in place of the small letters which represent the sides in the proportion, and the general form becomes so far adapted to a particular case in the solution of rightangled triangles; but if the angle be given in degrees, how are we to know its sine or cosine, for that is the quantity which enters into the formula; and how are we to know the numerical value of R? For the present the student must be satisfied with the reply, that he can find the numerical value of any trigonometrical line corresponding to an angle of any given number of degrees, in a table at the end of the work. This is TABLE XXIV.* of Natural Sines. The degrees for angles or arcs of every magnitude within the quadrant will be found at the top of the columns of the table, and the minutes in the column marked м on the left, if the given angle or arc be less than 45°; but if it be greater than 45°, the degrees will be found at the bottom of the page, and the minutes on the right; the length of the sine or cosine will be found in the column under or over the degrees, as the case may be, and on the same horizontal line with the minutes. The title of the column must be looked for at top if the arc be less than 45°, and at bottom if the arc be greater. The other trigonometrical lines may be easily calculated from the sines and cosines, as will be seen in the examples.

The trigonometrical lines of this table are computed, by a rule which will be hereafter demonstrated, for a circle whose radius is 1. So far as the principles for the solution of triangles are concerned, the length of the radius is entirely immaterial, as it will be recollected that the arc in the last diagram was described with any radius at pleasure.

When, in cases of the solution of right angled triangles, the hypothenuse and one of the acute angles are either given or required by the problem, one of the above formulas is always employed.

* The tables are selected and printed from the stereotype plates of a very large collection.

† A decimal point must be understood at the left of all the numbers in the columns of the table entitled N. sine and N. cos.