Example 5. Resolve into factors the expression

cos2a + cos2 ß+cos2y+2 cos a cos ẞ cosy - 1,

and shew that it vanishes if any one of the four angles a±ẞ±y is an odd multiple of two right angles.

The expression=cos2a + (cos2B+cos2y-1)+2 cos a cosẞ cos y

=cos2a+(cos2B - sin2y) +2 cos a cos ẞ cos y

α

= cos2a+cos (B+y) cos (ẞ − y) + cos a {cos (B+ y) + cos (B − y)} = = {cos a+cos (B+y)} {cos a+cos (ẞ − y)}

that is, if a±ẞ±y=(2n+1) π, where n is any integer.

Example 7. In any triangle, shew that

Za3 cos A=abc (1+4II cos 4).

By substituting these values in the given identity, and dividing by k3, we have to prove that

Now

Σ sin3 A cos A = sin A sin B sin C (1+41 cos A).

82 sin3 A cos A=42 sin2 A sin 24

=22 (1- cos 2A) sin 24

=22 sin 24-Σ sin 44;

and it has been shewn in Example 1, Art. 135, that Σ sin 24-41 sin A;

and it is easy to prove that

Σ sin 44 - 4II sin 24= -32ПI sin A. II cos A; .. 8 sin3 A cos A=8II sin A+ 32II sin A. II cos A; Σ sin3 A cos A = II sin A (1 + 4II cos 4).

EXAMPLES. XXIV. a.

1. If 0=a, and 0=ẞ satisfy the equation

If a and B are two different solutions of a cos 0+b sin 0=c, prove that

8. If a cos a+b sin a=a cos ẞ+b sin ß=c, prove that 2ab

sin (a+8)=

2ab
a2+b2,

and cot a+cot = c2 — a2 •

If cos 0+cosa and sin+sin p=b, prove that

1-cos2a-cos2 B-cos2y+2 cos a cos B cos y

as the product of four sines, and shew that it vanishes if any one of the four angles a±ẞ±y is zero or an even multiple of π.

14. Express

sin2a+sin2 B-sin2y+2 sin a sin ẞ cos y as the product of two sines and two cosines.

15. Express

sin2 a+sin2 ß+sin2 y − 2 sin a sin ß sin y − 1

as the product of four cosines.

In any triangle, shew that

19. Za3 sin B sin C-2abc (1+cos A cos B cos C').

20. Za cos3A =

21.

abc

4R2 (1-4 cos A cos B cos C').

a3 cos (B-C)=3abc.

22. If a and B are roots of the equation a cos 0+b sin 0=c, form the equations whose roots are

(1) sin a and sin ß; (2) cos 2a and cos 28.

Alternating Expressions.

305. An expression is said to be alternating with respect to certain of the letters it contains, if the sign of the expression but not its numerical value is altered when any pair of these letters are interchanged.

Thus

cos a-cos ẞ, sin (a-B), tan (a-B),

cos2 a sin (ẞ-y)+cos2 ß sin (y− a)+cos2 y sin (a− B)

are alternating expressions.

306. Alternating expressions may be abridged by means of the symbols and II. Thus

Σ sin2 a sin (8-y)=sin2 a sin (B− y) +sin2 ß sin (y − a)

+sin2 y sin (a-B);

II tan (6-7)=tan (ẞ − y) tan (y− a) tan (a− ß).

We shall confine our attention chiefly to alternating expressions involving the three letters a, ß, y, and we shall adopt the cyclical arrangement B-y, y-a, a- -B in which ẞ follows follows B, and a follows Y.

Example 1. Prove that Σ cos (a+0) sin (8 − y) =0.

Σcos (a+0) sin (8− y) =Σ (cos a cos 0 — sin a sin 0) sin (8 − y)

-

a,

=cos e Σ cos a sin (ẞ− y) − sin 0 Σ sin a sin (8 − y) =0,

since Σ cos a sin (6-7)=0 and 2 sin a sin (6-7)=0.

Example 2. Shew that Σ sin 2 (6 − y) = − 4II sin (8 − y).

sin 2 (B-) + sin 2 (y - a) + sin 2 (a - ẞ)

= 2 sin (ẞ − a) cos (a+ẞ- 2y)+2 sin (a – B) cos (a — B)

=2 sin (a - ẞ) {cos (a -B) - cos (a+ẞ-2y)}

=4 sin (a-8) sin (a) sin (8-)

=- 4II sin (8-y).

Example 3. Prove that

(1) Etan (6-7)= II tan (ẞ − y);

(2) Σ tan ẞ tan y tan (ß − y) = − II tan (8 − y).

(1) From Art. 118, if A+B+C=0, we see that
tan A+tan B+tan C=tan A tan B tan C.

Hence by writing A=8-y, B=y-a, C=a-ß, we have

Σtan (6-7)=II tan (8 − y).

(2) From the formulæ for tan (B− y), tan (y− a), tan (a -ß), we have

Σ (1+tan ẞ tany) tan (ẞ − y) = Σ (tan ß-tan y) =0;

whence by transposition

Σ tan ẞ tan y tan (ẞ-y) = -Σ tan (8-y)

= - II tan (ẞ-y).