By the aid of the formulæ of Art. 68 we may express the ratios of any angle in terms of any one of its ratios, e.g.

To express the other trigonometric ratios of in terms of its tangent.

1

tan 0

sec = √1 + tan20.

1

sec 0

1
FIG. 45.

cot =

cos 0:

=

M

=

1 sin

=

71. If be less than 90°, the following method can be employed with advantage:

Let MOP be a right triangle; Fig. 45, Let MP=t, OM=1, .. OP=√1+ť2. tan 0 = =t, sin e

=

=

sin 0: =

1

V1+tan20

V3 =

=

±√1+tan20
tan 0

cos 0 =

and so on. With suit

able modifications we may employ this method whatever be the magnitude of 0.

√3

√1+3

tan 0

Vtan20 +1

Hence if we are given tan we can calculate the value of the remaining ratios by either of the above methods. E.g. suppose tan 0 = √3, then

72. The expressions obtained for sin 0, cos 0, sec 0, and csc 0, in Art. 70, are affected with a double sign. Hence these ratios are not uniquely determined. If, however, we know tan ✪ and in addition to what quadrant belongs, we can determine the respective signs of the ratios named.

E.g. if tan √3 and we know that is of the third quadrant,

0

then sin

EXAMPLES. XV. (a).

1. Express all the other ratios of A in terms of cos A.

2. Express all the ratios of 90° + A in terms of cot A; A<90°.

3. Express all the ratios of 180°. A in terms of sec A.

4. Express all the other ratios of A in terms of csc A.

5. Use formulæ of Art. 67 to express all the other ratios of A in terms of sin A.

6. Use the method of Art. 71 to express all the ratios of A in terms of cosine A; A<90°.

EXAMPLES. XV. (b).

Suppose no angle in the following list is greater than + 180°.

1. If sin A = 3, find tan A and cosec A.

2. If cos B }, find sin B and cot B.

=

=

3. If tan A†, find sin A and sec A.
4. If sec = 4, find cot 0 and sin 0.
5. If tan 0 = √3, find sin ℗ and cos 0.

7. If sin @=

6. If cot = -, find sin @ and sec 0.
√5

find tan 0.

с

a, find sin 0 and cos 0.

= a, find sin @ and cot 0.

E

नेत

8. If tan

9. If sec

10. If sin 11. If cos

h, and tan 0=k, find the equation connecting h and k. 12. If A < 90°, and tan A + sec A = 2, prove that sin A = 3. Find the remaining ratios of A.

13. Show, using the formulæ Art. 67, that the numerical value of

sin A 1.

cos A 1.

sec A1.

cosec A1.

= a, and tan 6 = b, prove that (1 − a2) (1 + b2) = 1.

73. The formulæ developed in Chapter VII. and the expressions to be proved in Examples XIV. are true for any angle.

sin 0 cos

or

THE SOLUTION OF TRIGONOMETRICAL EQUATIONS

E.g. sin2 A+ cos2 A=1, whatever be the value of A. is true whatever value @ may have.

We shall now consider expressions which are true only for certain values of the angle.

=

E.g. sin V ¥24 is an equation satisfied by 45° and 135°, and by no other positive angles less than 360°.

The former expressions are called Trigonometric Identities. The latter expressions are called Trigonometric Equations. The solution of a trigonometric equation is the process of finding an angle which, if substituted in the equation, satisfies it.

or,

74. EXAMPLE 1. Solve the equation sin @ Substitute in the equation csc 0 =

Then

CHAPTER IX

II. Suppose
Then

No angle satisfies (1).

I. Let us suppose that Then from (2)

(sin

or,

or, in a single statement,

2 sin2 + 3 sin 0 - 2 = 0,
+ 2) (2 sin @ − 1) = 0.
.. sin 02,

sin 0.

csc0+= 0.
1
sin @

[Example 13, XV., (b).]
<180°.

0 = 30°,

0 = 150°.

is unrestricted in magnitude.

0 = 30°, 2 π + 30°, 390°, ... 2 nя + 30°;

0 = 150°, 2+150° ... 2 n + 150°,

0 = 2 nπ +30°, or (2 n + 1) π — 30°.

=

tan 0

(1) (2)

(Art. 41.) (Art. 58, Cor.)

(Art. 58, Cor.)

Another series of angles will satisfy equation (2),

- 330°, i.e. - 2π +30°, and in general - 2 n + 30°,

- 210°, i.e. -π 30°, and in general (2 n + 1) π - 30°,

viz.

and

n being a positive integer.

Another statement combines the results of II. thus,

0 = 2 nπ +30°,

0 = (2 n + 1) π — 30°,

n being any integer, positive or negative, or zero.

or

EXAMPLE 2. What positive angles satisfy the equation

√2 cos 0=cot @ ?

cos Ꮎ

sin 0

0,

Substitute

Then

1. sin @

√2 cos 0

=

cot 0 =

COS 8 sin @

From (1) 0

π

From (2) 0 2 nπ + or (2 n + 1) π

4

n being a positive integer or zero.

1

sin @

.. cos 0 = 0,
sin @=

1

V2

π
"

2

2π +

=

0.

3 п +

and in general në +

19.

=

EXAMPLES. XVI.

Find all positive angles not greater than 360° satisfying the equations: 1

V2

9. 2 cos @=
10. tan
11. tan

√3 cot 0.
3 cot 0.
cot 0 = 2.

2. 4 sin @csc 0.

3. 2 cos 0 sec 0.

12. 2 sin2 0

+ √2 cos 0 = 2.

4. 4 sin 0-3 csc 0 = 0.

= 0.

5. 4 cos 0 - 3 sec
6. 3 tan

cot 0.

13. 2 cos2 + √2 sin 0 = 2.
14. 3 tan2 0 - 4 sin2 0 = 1.
15. 2 sin20+ √2 sin 0 = 2.
16. cos20√3 cos 0 + } = 0.
17. cos2 + 2 sin2 0 — § sin 0 = 0.

7. 3 sin 0-2 cos2 0 = 0.
8. 2 sin 0 tan 0.

sin (180° - 0)

2

(Art. 58.) (Art. 58.)

kia

Find all angles that satisfy the following equations: 20. tan2 = 1. 21. √3 sin + 2 cos2 0 = 2.

2

Find all angles, either positive or negative, whose magnitude is not greater than 360°, which satisfy the following equations:

18. 5 tan20 sec20 11.

+tan 0 = 0.

(1)

(2)

ON THE TRIGONOMETRICAL RATIOS OF TWO OR MORE ANGLES

CHAPTER X

75. We will now establish the following fundamental formulæ :

sin (A + B) = sin A⋅ cos B + cos A· sin B

sin A. sin B

cos (A + B) = cos A⋅ cos B
sin (AB)= sin A⋅ cos B
cos (A-B)= cos A⋅ cos B

76. To prove that

and that

Here A and B are angles; so that (A + B) and (A — B) are also angles.

Hence, sin (A + B) is the sine of an angle, and must not be confounded with sin A+ sin B.

B
ΤΑ

Sin (A+B) is a single fraction.

Sin A+ sin B is the sum of two fractions.

The student should notice that the words of the two proofs of Arts. 76, 77 are very nearly the same.

H

F

sin (A + B) = sin A⋅ cos B + cos A · sin B,
cos (A+B)= cos A. cos B-sin A · sin B.

M

- cos A· sin B

N

K

+ sin A · sin B

FIG. 46.

(i.).

X