If we suppose the arc to be 0, the sine will be 0; as the arc increases, the sine increases until the arc becomes equal to 90°, when the sine becomes equal to +1, which is its greatest possible value; as the arc increases from 90°, the sine goes on diminishing until the arc becomes equal to 180°, when the sine becomes equal to +0; as the arc increases from 180°, the sine becomes negative, and goes on increasing numerically, but decreasing algebraically, until the arc becomes equal to 270°, when the sine becomes equal to -1, which is its least algebraical value; as the arc increases from 270°, the sine goes on decreasing numerically, but increasing algebraically, until the arc becomes 360°, when the sine becomes equal to — 0. It is 0, for this value of the arc, in accordance with the principle of limits.

+∞ to

The tangent is 0 when the arc is 0, and increases till the arc becomes 90°, when the tangent is; in passing through 90°, the tangent changes from 2, and as the arc increases the tangent decreases, numerically, but increases algebraically, till the arc becomes equal to 180°, when the tangent becomes equal to 0; from 180° to 270°, the tangent is again positive, and at 270° it becomes equal to +; from 270° to 360°, the negative, and at 360° it becomes equal to

tangent is again

- 0.

If we still suppose the arc to increase after reaching 360° the functions will again go through the same changes, that is, the functions of an arc are the same as the functions that are increased by 360°, 720° &c.

By discussing the limiting values of all the circular functions we are enabled to form the following table:

sented in the figure. Then we shall have, by definition

M

N

From the right-angled triangle OPM, we have,

PM2 + OP2 = 0M2,

The symbols sin'a, cos2a,

&c., denote the square of the
cosine of a, &c.

sine of a, the square of the
From Formula (1) we have, by transposition,

From the similar triangles ONM and OBT', we have,

Multiplying (6) and (7), member by member, we have,

From the similar triangles OPM and OAT, we have,

From the similar triangles ONM and OBT", and OBT", we have,

From the right-angled triangle OAT, we have,

0T = 01 + ᎪᎢ ; or, sec2a1+ tan2a.

(13.)

From the right-angled triangle OBT", we have, OT" OB2 + BT2; =

or, co-sec2a = 1 + cot2a. (14.)

It is to be observed that Formulas (5), (7), (12), and (14), may be deduced from Formulas (4), (6), (11), and (13), by substituting 90° a, for a, and then making the proper reductions..

ure, we shall discover the following relations, viz.:

FUNCTIONS OF ARCS FORMED BY ADDING AN ARC TO, OR SUBTRACTING IT FROM ANY NUMBER OF QUADRANTS.

63. Let a denote any arc less than 90°.

From what

Now, suppose that BM' = a, then will AM' = 90° + a

We see from the figure that,