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When the angle A is 0°, MP is zero, and when A is 90°, MP is equal to OP; and as A continuously increases from 0° to 90°, MP

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increases continuously from zero to OP; also OP is always equal to OX.

=

Therefore, when A 0°, the fraction. 0; when A

MP
OP

MP

OP

=

0 is equal to that is OP' 90°, the fraction is equal to that is 1; and as OP OP' A continuously increases from 0° to 90°, the numerator of the fraction MP continuously increases from zero to OP, while the denominator is OP unchanged, and therefore the fraction which is sin A, increases

MP

OP'

continuously from 0 to 1, and is positive.

As A increases from 0° to 90°, MP increases from zero to OP, and is positive.

Therefore sin A increases from 0 to 1, and is positive.

As A increases from 90° to 180°, MP decreases from OP to zero, and is positive.

Therefore sin A decreases from 1 to 0, and is positive.

As A increases from 180° to 270°, MP increases numerically from 0 to OP, and is negative; hence sin A increases numerically from 0 to 1, and is negative.

As A increases from 270° to 360°, MP decreases from OP to zero, and is negative.

Therefore sin A decreases numerically from 1 to 0, and is negative.

COROLLARY. Therefore we may conclude that sin A is never greater than 1; that cos A is never greater than 1. That the numerical value of sec A or of cosec A is never less than 1. For

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EXAMPLES. XIII.

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

1. cos A.

3. cot A.

5. cosec A.

7. sin2 A.

2. tan A.

4. sec A.

6. 1 — sin A.

8. sin Acos A.

ON THE RELATIONS BETWEEN THE TRIGONOMETRIC RATIOS

66.

P

ДА

-X

M

M

To prove

FIG. 41.

Similarly

CHAPTER VIII

O

FIG. 43.

cosec A

sec A

FIG. 44.

Let XOPA, any angle. (The figures are those of Art. 25.) The following relations are evident from the definitions:

cot A

sin A

tan A =

sin A COS 4

=

-X

1 sin A 1

COS A

1

tan A

sin A

cos A

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cos A
= cot A.
sin A

P

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Since

Similarly we may prove that

we have

=

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=

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MP2
Op2'

MP2 OM2
+
OP2 Op2

MP2+OM2 = OP2.

sin A tan A = cos A'

1 sin A

cos2 A

sin2 A+ cos2 A
cos Asin A

68. The following is a List of Formula with which the student must make himself familiar:

sec A=

1
tan A'
sin2 A+ cos2 A = 1,

1 + tan2 4 =

1+ cot2 A = cosec2 A.

=

tan A=

=

cot A =

OM2

OP

sec2 A,

=

1
cos Asin A

MP2 + OM2
OP2

tan Acot A =

1

cos A

sin A

cos A

tan2 4+1 sec2 A,

cot2 4+1 = cosec2 A.

Prove that tan A+ cot A sec A cosec A.

cos A
sin A

1
sec A =
cos A'

sin A cos A
+
cos Asin A

sec A cosec A.

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cot A

69. By means of these formulæ we are able to transform a given trigonometrical expression into a great variety of equivalent expressions.

EXAMPLE.

OP2

OP2

(Pythagorean Prop.)

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cosec A

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70. It is sometimes convenient to write a given expression in terms of the sine only, or in terms of the cosine only.

EXAMPLE I. Prove that sin+ 0 + 2 sin2 @ cos2 = 1 − cos1 0. By Art. 67, we have sin2 hence

= 1 cos2 0,

sin +2 sin2 0 cos2

= (1 - cos2 0)2+2 (1

cos20) x cos20

= (1 − 2 cos2 0 + cos1 0) + (2 cos2 0 - 2 cos1 0)
= 1 - cos1 0.

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EXAMPLE II. Express sin40+ cos 0 in terms of cos 0.

sin10+ cos 0 = (1 — cos2 0)2 + cos1 0

= (1 - 2 cos2 0 + cos10) + cos1 0

= 1 - 2 cos2 0 + 2 cos1 0.

NOTE. (1 cos 0) is called the versed sine of 0, and is written versin 0.

21.

Y 22.

Prove the following statements :

1. cos Atan A = sin A.

2. cot Atan A = 1.

3. cos Asin A. cot A.

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1

- sin A 1 + sin A

4.

5.

6.

=

7. (tan Acot A) sin A⋅ cos A
8. cos2 A - sin2 A = 2 cos2 A - 1
9. (sin A+ cos A)2 = 1 + 2 sin A⋅ cos A.
10. (sin A cos A)2 = 1 - 2 sin A⋅ cos A.
11. cost B

sin B2 cos2 B - 1.

19. ✓ 20. tan Atan B cot A+ cot B cot atan 8 tan a cot B

12. (sin2 B+ cos2 B)2 = 1.

13. (sin2 B cos2 B)2 = 1 - 4 cos2 B+ 4 cos1 B.

14. 1-tan4 B = 2 sec2 B – sec1 B.

Y 15. (sec B – tan B) (sec B + tan B) = 1.

16. (cosec - cot 0) (cosec 0 + cot 0) = 1.

17. sin30+ cos3 0 = (sin 0 + cos 0) (1 − sin @ cos 0).

-

V

18. cos3 @ sin3 0 = (cos sin 0) (1 + sin 0 cos 0).
sin60+ cos 0 = 1 − 3 sin2 0. cos2 0.
(sino 0 - cos 0) =
tan Atan B.

(2 sin2 0 — 1)(1 − sin2 0 + sin1 0).

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EXAMPLES. XIV.

-

(sec A.

=cot a tan ẞ.

-

28. (sin2 - cos2 )2.

29. 1 tan1 0.

tan A)2.

sec Acot A: = cosec A.

cosec A. tan A

= sec A. (tan A+ cot A) sin A cos A = 1.

·

sin2 A — cos2 A.

1 - 2 sin2 A.

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38. cos2 [(2 n + 1) π + A]

24.

Express in terms of (i.) cos e, (ii.) of sin 0,

27. cos40 sin1 0.

=

1 + cos A

1 cos A

versin20 sin2 0.

25. 2 versin ✓ 26. versin 0 (1 + cos 0) = sin2 0.

36. tan (180° + 45°) tan 45° + 1 = sec 245°.

37. cot 120° cot (60°) + 1 =

= (cosec A+ cot A)3.

31. tan2 + cot2 0.

32. 1+cot 0.

33. 1+cot2 0- cosec2 0.

34. 2 tan1 0 - 4 sin2 0.

30. sin60+ cos 0.

35. Show cos (90° — a) sin (180° — a) — sin (90° — a) cos (180° — a) = 1.

10$ AL

=

1
sin2 60° cos2 30°
cos2 A+ sin2 (1800 — A).

sec2 A

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