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1st. The sine is positive in the first and second quadrants, and

negative in the third and fourth : 2d. The cosine is positive in the first and fourth quadrants,

and negative in the second and third :

In other words, 1st. The sine of an angle less than 180° is positive ; and the

sine of an angle greater than 180° and less than 360°, is

negative : 2d. The cosine of an angle less than 90° is positive; the cosine

of an angle greater than 90°, and less than 270°, is negative; and the cosine of an angle greater than 270°, and less than 360°, is positive.

62. The algebraic signs of the sine and cosine being determined, the signs of all the other trigonometrical functions may be at once established by means of the formulas of Table I.

Thus, for example,

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Now, if the algebraic signs of sin a and cos a are alike, the tangent is positive ; if they are unlike, it is negative. Hence, the tangent is positive in the first and third quadrants, and negative in the second and fourth.

The same is also true of the cotangent: for,

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the sign of the secant is always the same as that of the cosine And since,

1 cosec a

sin a the sign of the cosecant is always the same as that of the +

+

I t +

64. The versed sine is constantly positive. For, it is always found by subtracting the cosine from radius, and the remainder is a positive quantity, since the cosine can never exceed radius. When the cosine is negative, the versed' sine becomes greater than radius.

65. Let q denote a quadrant: then the following table will show the algebraic signs of the trigonometrical lines in the different quadrants.

First q. Second

9. Third q. Fourth 9. sine

+
cosine
tangent

+
cotangent

+ 66. We have thus far supposed all angles to be estimated from the line CA from right to left, that is in the direction from A to B, to D, &c., and

B have also regarded such angles as posi. tive. It is sometimes convenient to give different signs to the angles them- D selves. If we suppose the angles to be esti

E mated from CA, in the direction from left to right, that is, in the direction from A to E, to D, &c., we must treat the angles themselves as negative, and affect them with the sign

For a negative angle less than 90°, the sine will be negative, and the cosine positive: for one greater than 90° and less than 180°, the sine and cosine will both be negative. The algebraic sign of the sine always changes, when we change the sign of the arc, but the sign of the cosine remains the same. Hence, calling x the arc, we have in general,

2 ) cos (-x) tan (-2) cot (– )

cot x. 67. We shall now examine the changes which take

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A

sin (

sin , cOS X, tan ,

place in the values of the trigonometrical lines, as the angle increases from 0 to 360°, and shall begin with the sine and cosine.

When the arc is zero, the sine is 0, and the cosine equal to R= 1. At 90° the sine becomes equal to R= 1, and the cosine becomes 0. At 180°, the sine becomes 0, and the cosine equal to R 1. At 270°, the sine becomes equal to R =

- 1, and the cosine equal to 0. At 360°, the sine becomes equal to 0, and the cosine to R = 1. Hence,

First quadrant.
As the arc increases from 0 to 90° :
The sine increases from 0 to 1:
The cosine decreases from 1 to 0.

Second quadrant
As the arc increases from 90° to 180° :
The sine decreases from 1 to 0:
The cosine increases, numerically, from 0 to - 1.

Third quadrant.
As the arc increases from 180° to 270° :
The sine increases, numerically, from 0 to – 1:
The cosine decreases, numerically, from - 1 to 0.

Fourth quadrant.
As the arc increases from 270° to 360°:
The sine decreases, numerically, from – 1 to 0:
The cosine increases from 0 to R

= 1.

68. By a careful consideration of the preceding principles and by making the proper substitutions in the formulas already deduced, we may now form the following Table:

TABLE II.

sin a,

COS Oliy

tana,

sin din cota,

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COS

sin 0

0, sin (180° + a) COS 0

1, cos (180° + a) tan 0

: 0,

tan (180° + a) cot 0

= 0.
cot (180° + a) =

cot a. sin (90° a)

= COS A,

sin (270° a) cosa, cos (90° a) = sin a,

cos (270° a) tan (90° a) = cot a, tan (270° – a) cot (90° a) = tan a. cot (270° a)

tan a. sin 90°

1,
sin 270°

1,
90° = 0,

cos 270°

0, tan 90°

tan 270° cot 90°

0.
cot 270°

0. sin (90° + a) cos a, sin (270° + a)

COS a, cos (90° + a)

cos (270° + a) tan (90° + a) cot a, tan (270° + a) cot (90° + a) tan a. cot (270° + a) tan a. sin (180° – a) sin a, sin (360°

sin (360° – a) cos (180° a) cos a, cos (360° a)

COS di, tan (180° a) tan a, tan (360° a) cot (180° – a)

cot (360° a)

cot a. sin 180° 0, sin 360°

0,
1,
cos 360°

1,
tan 180°
0, -tan 360°

= 0, cot 180°

cot 360°

sin ag

sin d, cota,

sin a,

tan a,

cot a,

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cOS 180°

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8

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69. The examinations thus far, have been limited to arcs which do not exceed 360°. It is easily shown, however, that the addition of 360° to any arc as x, will make no difference in its trigonometrical functions; for, such addition would terminate the arc at precisely the same point of the circumference. Hence, if C represent an entire circumference, or 360°, and n any. whole number, we shall have,

sin (C + x) = sin x; or, sin (n X C + x) The same is also true of the other functions.

- sip x.

70. It will further appear, that whatever be the value of an arc denoted by x, the sine may be expressed by that of an arc less than 180°. For, in the first place, we may subtract 360° from the arc 2, as often as 360° is contained in it: then denoting the remainder by y, we bave,

sin x = sin y. Then, if y is greater than 180°, make

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Thus, all the cases are reduced to that in which the arc whose functions we take, is less than 180°; and since we also know that,

sin (90 + x) = sin (90 – ), they are ultimately reducible to the case of ares between 0 and 90°.

GENERAL FORMULAS.

71. To find the formula for the sine of the difference of two

angles or arcs.

Let ACB be a triangle. From the vertex C let fall the perpendicular CD, on the base AB, produced.

Denote the exterior angle CBD by А B D a, and the angle CAB by b. Then, AB= AD DB. But (Art. 25), AD = AC cos b, and BD = BO cos CBD. Hence, AB AC cos b - BC cos a. Dividing both members by AB, we have

AC

BC
1 =
AB

AB
But, since sin a= sin CBA, we have (Art. 21)
AC

cos 6

COS a.

BC sin b

and AB sin ci

AB sino

sin a

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