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Then, in the three cases, we take (BC), (CA), (AB) as primarily positive.

And hence (LA), (MB), (NC)—which make positive rightangles with these—are secondarily positive.

But the three sets of conventions must not otherwise be combined.

For, as we see in Art. 322, when we drop a perpendicular on (BC), (BA) and (AC) (which follow the order B, C) are regarded as positive: whereas they are each negative, when taken by themselves as initial lines.

325. The convention, at which we have arrived with respect to perpendiculars on the sides, is of great importance. It is as follows:

Perpendiculars drawn from a side towards the opposite angle are positive.

Example 1. If J be any point, and if x, y, z represent perpendiculars from J to the sides—directed from the sides to J—and a, b, c represent the lines (BC), (CA), (AB), and if a represent the area (ABCA); then in every case

ax+by+cz=20, This follows from Ex. x. 110.

Example 2. Let O be the orthocentre of ABC; and AL perpendicular upon BC. Apply the convention of signs to the algebraical expressions for LO and A.

If ABC is an acute-angled triangle we see from Art. 212, that LO=2R cos B cos C; and OA=2R cos A.

Now, if A is obtuse, OA must be drawn towards BC; because O is within the space between BA and CA produced. But in this case cos A is negative. Hence we have always

(OA) directed=2R cos A. Again, if B is obtuse, (LO) must be drawn away from A, because O is within the space between CB and AB produced. But in this case cos B is negative. Hence we have always

(LO) directed=2R cos B cos C. Furthermore,

(LO)+(0A)=(LA). Now 2R cos B cos C+2R cos A=2R cos B cos C-2R cos (B+C)

=2R sin B sin C,

=b sin C, which is the obvious value of (LA). J. T.

16

EXAMPLES XII.

1. If A, B, C, D, E are points taken in order on a line, find the directed length which is equivalent to each of the following combinations of directed lengths :(1) (AB) + (BC) + (CD)+(DE). (2) (AC)+(CE) (DE). (3) (BE) + (ED) (CD). (4) (DE) + (EA) (DC). (5)(CA)+(AE)-(DE)+(DB). (6)(AD)+(DB)+(BE)+(EC)+(CA).

2. Show how symbols denoting directed lengths may be used in solving the following problem :

Two persons A and B walk along a road in the same direction :-A, at the rate of a miles an hour; B, at the rate of b miles an hour. At a certain moment A is at a place c miles from a town 0 (towards which he is then walking), and B at a place d miles from the same town on the same side of it. Find where and when A and B are to be found together : distinguishing the cases according as a > or <b, c> or <d, a/c> or <b/d.

Prove the following identities : -(3—8).
3. (sin 150° – cos 150°) (sin 120° + cos 120°) = cos (- 60°).
4. (sin 225o + sin 240°) (cos 210° cos 225°) = sino 150°.
5. sin 210° + sin 240° = /2 cos 165o.
6. cos (– 30°)+cos 120° = /2 sin 165°.
7. cos (- 36°)= cos 300° + cos 288°.
8. sec 37 = tan 855°

= cosec 177 = 2 cos 840°. 9. Given sec A = 7, find tan A and cosec A. Explain the reason why each result is ambiguous.

10. Given 180° > A > 90°, and sin A = }; find cos A.
11. Given 270o <A < 360°, and tano A = £, find sin A.

12. Give the complete solutions of the following equations :-(13—32).

1

13 13. sin A = 1. 14. tan A

15. CoS A =

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16. tan A = tan 19°. 17. 2 sino A = 1. 18. 12.sin 4 = 1. 19. sinë cos0. 20. sin e

= cos 0.

21. tano 0 = 3. 22. 2 cos x + 2 sec x= 5. 23. 3 tan? 6 + 8 cos” 0 = 7.

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28.

cot O – tan 0 = cos 0 + sin 8. 29. sec 0 - 2 tano 0 = 2.

=+1.

30. tan po = cotq0, 31. vers 0 = 0. 32. tan o 33. Find the values of the ratios of the following angles :4п

277 250°, 765°, -60°,

° 3

3

8

'

150°,

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34. Trace the changes in magnitude and sign of each of the ratios of an angle which increases from 0 to 360°.

35. Show that a ratio always changes its sign, as it passes through the value 0 or o.

36. Show that for any value of A, sin (90° – A) = cos A ; cos (90° – A)=sin A ; sin (90° + A) = cos A; cos (90° + A) =- sin A.

37. Show that

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38. Trace the changes in sign and magnitude as A increases from 0 to 360° of

sin A + cos A ; sin A -cos A ; tan A + cot A ; tan A – cot A; sin A. cos A ; tan A . sec A ; cos A.cot A ;

cOS A + sin A

COS A - sin A [The values of these expressions at the beginning and end of each octant should be given.]

39. Prove

(i) If n is even, sin n (# – ) =(-1) 3- sin ne ;
(ii) If n is even, cos n (3# - ) = (– 1)cos no ;

n-1

n-1

a = cos-1

+

(iii) If n is odd, sin n (35 - 6) =(-1) 2 cos no;
(iv) If n is odd, cos n (4+ - 0) = (-1) ? sin no.

– 6( 40. Show that 1 (4n + 1) (1 - A) and NT + (-1)" A represent the same series of angles. 41. Show that (i) sin

(1 aạ) = NT, (ii) sec-1 a + tan-? (a? 1) = NT,

(iii) cot-- a = cosec-? (a2 + 1)=nt, if one or other of the signs is taken in each case. 42. If we interpret ym to mean the positive quantity whose

and sin-'a to mean the numerically smallest angle, whose sign is the same as that of a, and whose sine = a; and so cos-1 a, tan- a &c. : show that

(i) Ja?. sin-?a=a.cos-- J(1 a?),
(ii) Va. tan-lara. sec- [(a? + 1),
(iii) Va’.cos-la= Va’.sin-(1 - aʼ) + 3+ (a - Ja),
(iv) Ja”. sec-?a= Ja. tan- Ja’ – 1) + žo (a - Ja),
(v) Vas. sin-la+a.cos-lart. Ja?.

square is

m;

CHAPTER XIII.

RATIOS OF COMPOUND ANGLES.

§ 1.

SUM AND DIFFERENCE OF TWO ANGLES.

326. To find the sine and cosine of A + 90° and of A – 90°, for all values of A, in terms of the ratios of A.

Let OI be an initial, and OF a final line.

Let OK make a positive right-angle with 01, so that angle IOK is a positive right-angle.

Let IO be produced to I', so that KOI' is a positive rightangle. Then in determining the ratios of IOF,

OI is the direction for positive bases,

OK is the direction for positive perps.
But in determining the ratios of KOF,

OK is the direction for positive bases,

Ol' is the direction for positive perps.
Hence, for any given position of OF,

base for IOF= - perp. for KOF,
perp. for IOF= + base for KOF.
.. cos IOF= - sin KOF

.(1),
and sin IOF= + cos KOF

(2).

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