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=

cos a,

cot a,

cosec a,

=

cos a, cot a,

cosec a,

7. Let n be 4, and a be negative.

sin (360°-a)=sin a,

tan (360°— a) tan a,
= ―
sec (360°— a) =

sec a,

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COS

(270°— a) =
(270°-a)=
cosec (270°-- a) =

cot

=

8. Let n be 4, and a be positive.

sin (360° a) = sin a,
+
tan (360° a) tan a,
sec (360° a) = sec a,

=

(270° + a)=
(270° + a)
cosec (270°+a)

COS
cot

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=

COS

=

(360°— a) = cot (360° — a) cosec (360° — a) :

cos a,

cot a,

Cosec a.

=

=

sin a, tan a,

sec a.

sin a, tan a,

sec a.

cos α, cot a,

cosec a.

COS

(360°+ a) = cos α, (360°+ a) = cot a, cosec (360°+a):

cot

cosec a.

It will be observed that when is even, the functions in the two members of the equations have the same name; and that when n is odd, they have contrary names. The algebraic sign attributed to the second member is determined by the quadrant in which the arc is situated.

Let this article be reviewed, and these principles applied in determining the names and algebraic signs of the second members.

Hence, functions of arcs greater than 90° can be found in terms of functions of arcs less than 90°. Thus,

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If n is integral and positive, prove the following:

4. sin [n 180° + (− 1)" a]

sin a.

5.

cos (n 360° ± a) = cos a.

6. tan (n 180° + a) tan a.

7. Any function of (n 360° + a) = the same function of a, whatever be the value of a.

87. Values of Functions of Particular Arcs.

1. To find the functions of 30°.

Since 60° is one-sixth of the circumference, the chord of 60° is equal to one side of a regular inscribed hexagon, which is equal to the radius or 1. But the sine of 30° is equal to one-half the chord of 60°.

1

13

2

13

.. (1) sin 30°, ... (2) cos 30°=√1 — += V3.

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cos 30°.

= sin 20°.

- tan 15°.

Dividing (1) by (2), then (2) by (1), taking the reciprocals of (2) and (1), we have

(3) tan 30°

(4) cot 30°= √ 3.

(5) sec 30°

(6) cosec 30°

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From article 40, sin 60°: cos 60°

cos (90' — 30°)

=

2. To find the functions of 60°.

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2.

sin (90° — 30°) = cos 30°, sin 30. Hence,

1.

(1) sin 60°1/ 3,

(3) tan 60° 3,

=

2.

(5) sec 60° 2,

=

(2) cos

(4) cot

=

2 sin2 45° 1,
(1) sin 45√2,
(3) tan 45° 1,

=

(5) sec 45° = √2,

60°

60'

(6) cosec 60°

3. To find the functions of 45°.

From Art. 40, sin 45° sin (90° 45°) = cos 45°; but sin2 45+ cos2 45° = 1,

Prove the following:

sec 120° = - 2.

cos 135°12.

3. sin 30041/3.

=

1.

4. tan 225° :
9. Construct an angle whose tangent is
10. Construct an angle whose sine is — 1⁄2,
11. Find all the functions of 150'.

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.. sin2 45°. Hence,

(2) cos 45° = {√2,

(4) cot 45° 1,

(6) cosec 45° = √ 2.

cosec 210°-

5.

6. cot 240'

7. sin 390° 11/1/

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1

V 3

2

√3

ང་

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8. cos (120°)——

- 2.

1

V 3

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88. Inverse Trigonometric Functions.

==

If x sin a, then a is the angle or arc whose sine is x, which is written asin-1 x, and read a equals the arc whose ine is x.

It must not be supposed that -1 is an exponent, and

1

sin x

-1

that sin x

=

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; this would be a grievous error.

Let the following be read:

cos-1x, tan-1x, sec-1x, cosec-1x, sin-1 (cos x), sin (sin-1x),

1

1 sin-1 cosec 1 cos-1x

tan-1x

X

sec

If n0, (1) becomes,

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(1) ƒTM ƒ” (x) = ƒm+n (x).

... (3) ƒ° (x) = x.

The above notation is not altogether arbitrary; for let f(x) be any function of x, and let ƒ[ƒ(x)], or, more briefly, let f2(x) be the same function of f(x), which notation denotes, not the square of f(x), that is, not [f(x)], but that the same function is taken of f(x) as of x. Thus, if f(x) sin x, ƒ[f(x)] = sin (sin x), then, in general,

(2) fm ƒ° (x) = ƒTM (x).

cot=

-1

1

X

If m1, and n-1, (1) becomes,

=

(4) ƒƒ-1(x)=ƒ° (x) = x.

Hence, f(x) denotes a quantity whose like function is x.

Hence, if y = sin-1x, sin y sin (sin x) = x; that is, y or sin 1x is an arc whose sine is x.

-1

It would follow from the above that sin2 a ought to signify sin (sin a), and not (sin a)2; but since we rarely have sin (sin a), it is customary to write sin2 a for (sin a), as we are thus saved the trouble of writing the parenthesis.

It would not, of course, do to write sin a2 for (sin a)2, for then we should have the sine of the square of an arc for the square of the sine of an arc.

Let the following equations be proved:

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89. Problem.

To find the sine and co-sine of the sum of two angles.

Let a = the angle OCA, and b = the angle ACB. Draw BL perpendicular to CA, BP and LM perpendicular to CO, and LN parallel to CO.

=

=

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0

The triangles NBL and MCL are similar, since their sides are respectively perpendicular; hence, the angle NBL opposite the side NL equals the angle MCL opposite the homologous side ML. But MCL a; hence NBL : =a.

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B

C

From the diagram we find the following relations: (1) LB sin b. (2) CL = cos b. (3) PB ML + NB. (4) PB sin OCB sin (a + b). (5) ML sin MCL CL (6) NB = cos NBL × LB

sin a cos b.

cos a sin b.

P M

A

Substituting the values of PB, ML, and NB, found in (4), (5), and (6), in (3), and denoting the formula by (a), we have

sin a cos b + cos a sin b.

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