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94. The connection between the ancient and modern nomenclature is thus indicated :

If we divide each linear function of the arc by the radius of the circle, we obtain the corresponding ratio-function of the angle. Thus linear sine

MB

sine of angle AOB. radius

OB linear tangent

AT radius

OA

tangent of angle AOB.
linear secant OT

secant of angle AOB.
radius
linear versed sine MA

= versed sine of angle AOB.
radius

ОА linear subtense BT

- subtense of angle AOB. radius

OB

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=

ОА

95. Now, as the angle AOB or the arc AB increases, it is clear that each of the lines MB, AT, OT, AM, BT increases. Hence the primary ratios sine, tangent, secant, versed sine, subtense increase as the angle increases.

The functions, having the prefix co-, are found from the above by substituting for the arc or angle its complement. Hence the secondary ratios decrease as the angle increases.

96. Any relations which hold amongst the linear functions will involve the radius of the circle. The corresponding relations amongst the ratio-functions may be found from these by putting radius=1. Conversely, we may find the relations amongst the linear functions, from those involving the ratio-functions, by introducing radius wherever it is necessary to make the equations homogeneous. If we place rad. instead of 1 in the centre of the ratio-hexagon, we shall see the symmetry of the relations more clearly than before. Thus, connecting the linear functions, we have sin? + cosa=rada, tano+rada=seca, rad? +cot2 =cox?

sin , sec=rad, tan &c., cos . sec=rad? &c. sina. tano+sin?. rada=tana. rad?, cosa , rad? + cos2. cot=cot?. rad?, rada. seca+rada. cox2=seca. cox2.

.

.

EXAMPLES III.

:

1. Let ABC be a triangle, right-angled at C, such that AB=6 ft., BC = 711 ft. Find the trigonometrical ratios of the angles at A and B.

2. The sides of a right-angled triangle are in the ratio 3:4:5. Find sine, cosine, and tangent, of each of the acute angles.

3. The sides of a right-angled triangle are in the ratio 2:3J5 :7. Find secant, cosecant, and cotangent, of each of acute angles.

4. Write down five independent formulæ by means of which the other ratios may be expressed immediately in terms of sec A and tan A, and these in terms of one another.

5. Write down similar formulæ for cosec A and cot A.
6. Why is sin 30o = cos 60° ?
7. Show that sin 45o = cos 45°, and thence find tan 45o.
8. Apply the equation sino A + cos” A = 1, to find sin 45°.
9. Find the angle whose cosine is equal to sin 70°.

10. In a triangle, right-angled at C, the sides opposite A, B, and C, are called a, b, and c, respectively.

C, Prove (i) a sin B=b sin A.

(ii) 2a’ tan B + b2 sec? A =co (sin A + sin B)?. 11. Prove tan? A

cot? A (i) sin? A

(ii) cos’ A = 1 + tan’ A

1 + cot? A

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sec A

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S

13. Show from definition and figure

(i) cosec A - sin A = cos A. cot A. (ii)

cos A = sin A. tan A. (iii) secA + coseco A = seco A. cosec? A. (iv) tano A – sino A = tano A . sino A.

(v) cot? A – cosA = cot? A . cosA. Prove the following identities :

sec A-1 tan A 14.

tan A sec A + 1 1 + tan A

cot A +1 15.

1-tan A cot A-1' 16. 1 + tan 0 cosa 0 + cot & sino 0 = (sin 0 + cos 6)?

1 + cos 0 1 17.

- 4 cot 0. cosec 0. 1- cos 0

1 + cos

sin A 18.

= sec A - tan A. 1 + sin A

sin A 1- COS A 19. 1 + cos A

sin A 1 -cos A 20.

(cosec A

cot A)? 1 + cos A 21. sec A + sino A + cot’ A = coseco A + coso A (sec: A – 1). 22. sec 0 = sec 0 + tan” A. cosec 0.

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23.

+

24.

+

sec

1

1 2 sec? 0

+

1-sin 0 1+ sin 0°
cos 0 + sin 0

cos 0 – sin
cos 0 - sin cos 0 + sin 0
sec 0 cosec 0 cosec 0 + sec 0

cosec A
2 cosec 0 - sec 0 cosec 0 + sec
(sec A .cot À + 1) (sec A .cot A - 1) = cos’ A. cosec? A.
(sec? A + tan’ A)(coseco A + cot? A) = 1 + 2 sec? A . coseco A.
cott A + cot? A = cosec' A - coseco A.

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25.

26.

=

27.

28.

sec4 A - sec? A =tan: A + tano A.

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39.

=

29. (seco A. coseco A – tano A)(seco A.cosec? A – cot? A)

= 1 + 2 sec? A cosec- A. 30. (cosA + cot? A) tanA = seco A + (cosA – 1) tanA. 31. cos4 A - sin* A = cos? A - sino A. 32. cos* A + sin" A=1- 2 sin? A . cos? A. 33. cos A + sin A=1 - 3 sin? A. cos? A. 34. cos® A – sin® A = (coso A – sin? A)(1 - 2 sino A.cosA). 35. sec6 A=1 + tan6 A + 3 tano 0 . seco 0. 36. vers’ A - 2 vers A + sino A=0. 37. tan A - sin A tan A vers A. 38. vers A = cos A subt A.

vers A (1 + sec A) = subt A (1 + cos A). 40. sino A - sino B = cos B – cosé A. 41. tano A + seco B=seco A + tano B. cos A + cos B

cos B - cos A 42.

COS A.cos B sec A + sec B

sec A - sec B 43. (sin A + sin B) (cosec B - cosec A) = (sin A – sin B)

(cosec B + cosec A). sin A + sin B cos B - cos A 44.

cos A + cos B sin A - sin B 45. cos B) (sin A + cos B) = (sin B - cos A)

(sin B + cos A). 46. Solve for sin 0 :

(i) 3 sino 0 cos2 0 = 6 cot? .
(ii) sin 0 + cos 0 = 1.
(iii) tan 0 + sec 0 = 2.
(iv) cot 0 + cosec o 3.

(v) 4 cos? 6 + 2 sin 6 = 4 + 13 (1 – 2 sin 6). 47. Solve for cos 0:-cot A + cosec 0 = 5.

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49. In the ratio-hexagon, the ratio of the product to the sum of the squares of any two horizontally adjacent quantities is equal to the square of the quantity just above and between them.

CHAPTER IV.

RATIOS OF PARTICULAR ANGLES.

§ 1. ONE RATIO BEING GIVEN. 97. Given the value of any one ratio of an angle, to draw the angle and to find the values of the other ratios.

General Method. Let p:q be the given value of the ratio. Construct a right-angled triangle, in which the measures of those two sides, by which the ratio is defined, are respectively p and q. Then find the third side by Euc. I. 47.

98. Given the value of the sine or cosine of any angle, to draw the angle.

Let the given value be r, i.e. r : 1.

Then, since sin and cos are always less than 1, r is ess than 1.

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