57. sin (45° + A) — sin (45° – A) = √2 . sin A.

59. If tan A

tan (A-B).

60. If tan A =

=

2

and tan B = 9, find tan (A + B) and

and tan B = 1, show that tan (A + B) = 1. 61. If (1+tan A) (1 + tan B) = 2, show that tan (A + B) = 1. 62. Show that if 45° is divided into two parts, 0 and 6, then (1+tan 0) (1 + tan 4) = 2.

tan A tan 24 tan 3A

= tan 34 - tan 24 - tan A.

=

63. 64. If tan 0=2m+1 and tan = 2m-1, then cot (0−4) = 2m2. 65. sin (A + B). sin (4 – B) = sin2A – sin2B = cos2B – cos2A. 66. cos (A + B). cos (A - B) = cos2A - sin B = cos2B - sin2A. cos 2A

67. tan (45° - A)

Solve:

68. sin 20= tan 0.

69. sin = cos 20.

71. sin 0 sin 20.

=

74.

tan 0 + tan 20 + √3 tan 0. tan 20 = √3.

75. √3 (tan 0+tan 20) + tan 6. tan 20 = 1.

CHAPTER VI.

TRIGONOMETRICAL FORMULE FOR A TRIANGLE.

§ 1. THE TRIGONOMETRICAL RATIOS OF OBTUSE ANGLES.

136. We have at present defined the ratios of acute angles only. We now proceed to define those of obtuse angles.

In OI take any point B: and draw BH at right angles

to Ol' meeting OF in H.

Then BOH is the right-angled triangle by which we determine the ratios of the acute angle I'OF, OB being base, OH being hypothenuse, and BH being perpendicular.

Now it is found convenient to use this same right-angled triangle BOH to define the ratios of the obtuse angle IOF; only that, as OB has to be drawn in the opposite direction to the line OI, which bounds the obtuse angle, we take - OB (instead of OB) as its base.

Thus

Thus, those ratios of an obtuse angle, which contain the base - OB (viz. all but the sine and cosecant), are negative.

A fuller explanation of this adoption of the minus sign will be given later. But the student will immediately perceive its convenience, for by its means the formulæ connecting the ratios of the angles and the sides of a triangle are the same whether the triangle be acute-angled, right-angled or obtuseangled.

137. To find the ratios of the obtuse angles 180° - A and 90° + A in terms of those of the acute angle A.

Take the figure of last article.

H

I.

B.

I'

(1) Let BOH = A.

Then since IOF + BOH = 180°, .. IOF 180° – A,

BH

sin IOF= = sin BOH, i.e. sin (180° – A) = sin A,

он

Then since 10F = OBH + BHO, .. IOF=90° + A,

138. To find the ratios of the acute angles 180° – A and A – 90° in terms of those of the obtuse angle A.

Take the same figure.

(3) Let IOF = A, then BOH = 180° – A,

= sin IOF, i.e. sin (180° - A) = sin A,

BH

sin BOH

=

он

Example. To trace the changes in the ratios of obtuse angles.
As A increases from 90° to 180°,

tan A is -ve;

∞ to 0.

sin A is+TM; and algebraically decreases from 1 to 0..
cos A is; and algebraically decreases from 0 to -1.
and algebraically increases from
and algebraically decreases from 0 to -8.
sec A is -ve; and algebraically increases from ∞ to -1.

cot A is -";

cox A is+ve; and algebraically increases from 1 to ∞.

140. From the proposition that the angles of a triangle are together equal to two right angles many important results follow. Thus, if A, B, C are the angles of a triangle, we have

1.

2.

EXAMPLES VI. A.

Write down the ratios of 135°, 120°, 150°, 112, 105°.

Show that sin 120° = 2 sin 60° cos 60° and that

cos 120° = cos2 60° - sin2 60°.

3.

Show that tan 150° = cot 60° - cosec 60° and that
cos2 135° = cos 60°.

4. If A, B, C are the angles of a triangle, show that sin 1⁄2 (B + C) = cos 1⁄2 4; and that tan 1⁄2 (A + B) = cot 1 C.

5. Show that the ratios of the half-angles of a triangle are all positive.

6. Show that, given the sine or the cosecant of an angle of a triangle, that angle may have either of two values.

7. Show that the area of a triangle is measured by half the product of two sides and the sine of the internal angle contained by them, whether that angle is acute, obtuse or right.