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CHAPTER X**,

ON ANGLES UNLIMITED IN MAGNITUDE.

137. Just as the definitions of the Trigonometrical Ratios apply to angles of any magnitude whatever, so every general Formula involving these Ratios is true for angles of any magnitude whatever.

It is most important that the student should examine for himself into the truth of this statement.

sin A'

COS A:

138. The formula
1
1

1
cosec A

sec A=

cot A =

tan A' are really definitions; and since the definitions apply, therefore these formulæ are true, whatever be the magnitude of A.

sin A The formulæ tan A

cotA:

cos A
sin A'

cos A'

are deduced immediately from the definitions, and therefore they are true whatever be the magnitude of A. 139. The formulæ sino A + cos* A = 1,

1 + tan’ A = sec A,

1 + cot' A = cosec® A, are each a trigonometrical statement of Euc. 1. 47, and depend only on the fact that MP, OM, and OP are the sides of a right-angled triangle. That this is the case, whatever be the magnitude of the angle A, is evident from the figures on page 96.

** To be omitted on a first reading, except pp. 104, 105.

140. In Art. 118 we proved that the sine of an angle is equal to the cosine of its complement, provided the angle lies between 0° and 90°.

We now give some examples of a method of proving the truth of this and other like formulæ, whatever be the magnitude of the angle concerned.

Example 1. To prove that the sine of an angle=the cosine of its complement.

That is, to prove sin A=cos (90° — A) and

cos A=sin (90° – A). We take two revolving lines OP and OP. OP starting from OR is to describe the angle A; OP' starting from OR is to describe (90o – A).

As usual, PM, P'M' are perpendiculars on OR and P'N' is a perpendicular on OU.

In describing (90° – A) we shall consider that OP' starting from OR, turns first through 90° into the position OU, and then turns back from OU through the angle UOP'=(-A).

lu

p'

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D So that ROP, the angle which OP describes from OR, is always equal to UOP', the angle which OP' describes from OV in the opposite direction.

Hence, N'P', that is OM', is always equal to MP in magnitude.

Also it will be seen that when P is above LOR, P is to the right of UOD; when P is below LOR, P' is to the left of UOD. Hence, OM' and MP have always the same sign.

MP ОМ? Therefore

OP

op always,

or;

sin A=cos (909 – A), for all values of A.

Again, ON', that is M'P', is always equal to OM in magnitude.

And P' is above or below LOR according as P is to the right or to the left of UOD.

So that M' P' and OM have always the same sign.

Therefore

OM M'P
OP

always,
OP

or,

cos A=sin (90° – A) for all values of A.

EXAMPLES. XXVIII.

Prove, drawing a separate figure for each example, that (1) sin 30o=cos 600.

(2) sin 65o=cos 250. (3) sin 1950=cos ( - 1050). (4) cos 275=sin(-1859). (5) cos ( – 270)=sin 1170. (6) cos 300°=sin(-210).

If A, B, C be the angles of a triangle, so that A+B+C=1800, prove A B+C

B A+C (7) cos sin

(8) cos =sin 2

2

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(9) sin

A+B
COS

2

(10) sin =

А B+C
=COS

2

141. Def. Two angles are said to be the Supplements the one of the other when their sum is two right angles.

Thus (180° – A) is the supplement of A

If A, B, C be the angles of a triangle, (A+B+C)=180°, so that (B+C) is the supplement of A.

Example 2. To prove that the sine of an angle=the sine of its supplement; and that the cosine of an angle = -(the cosine of its supplement).

That is, to prove sin A=sin (180°– A) and

cos A= -cos (180o – A). We take two revolving lines OP and OP'. OP starting from OR describes the angle A; OP' starting from OR describes the angle (1800 - A).

In describing (180° – A) we consider that OP' starting from OR turns first through 180° into the position OL, and then back from OL through the angle LOP=(-A).

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So that ROP, the angle which OP describes from OR, is always equal to LOP', the angle which OP' describes from OL in the opposite direction.

Hence, MP and M'P' are always equal in magnitude.
Also, P and P are always both above, or both below LOR.
So that MP and M'P' are always of the same sign.

Therefore

MP
OP

M'P

always, OP

or,

sin A=sin (180° – A), for all values of A.

Again, OM and Ol' are always equal in magnitude.

Also it will be seen that when P is on the right of UOD, P' is on the left of UOD; when P is on the left of UOD, P' is on right of UOD.

So that OM and OM' are always of opposite sign.

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Prove, drawing a separate figure in each case, that

(1) sin 600=sin 1200. (2) sin 340°=sın ( - 160°). (3) sin(-40)=sin 2200. (4) cos 3200=

= - cos ( - 1409). (5) cos ( - 3800)= - cos 560°. (6) cos 195o = - cos ( - 15°).

If A, B, C be the angles of a triangle, prove
(1) sin A=sin(B+C). (2) sin C=sin (A + B).
(3) cos B= cos (A +C). (4) cos A= -cos (C + B).

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