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.. each of the angles FAB, FBA of the triangle FAB is double of the third angle AFB. :: <AFB+ 2.. AFB+2.. AFB = 180°, i.e. 5.4 AFB= 180°,

.. AFB= 36o. Join FC. Then FC bisects angle AFB and is perpendicular to AB.

AF .. LAFC = 18°, and cosec AFC =

AC Now let AC =CB=1. Then AB= BD = 2, and CEP = C DP=C B2 + BDP = 1+ 22 = 5, :. CE=5. and AF= BF= AE=CE + AC = 15+1.

... cosec 18° = 15+ 1. The following construction gives the ratios of 18° or 72° and of 54° or 36° in their simplest form.

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In the above figure, with centre 0 and radius OF or OA describe a circle FAGE cutting AB produced in E, and FB produced in G. From F and G drop FC, GH perpendiculars on AE. These perpendiculars being drawn from the extremities F, G of the diameter FG cut off equal segments from the chord A E.

.. HE=AC and AH=CE. Consider the angles of the figure. :: FAG=90° and FAB=72°, .. BAG=18°, .:: GFE=GAE=18°. AEG=AFG =36o. FGE=FAE=72°, :: FEG=90° and A EG=36°, .. AEF=54°=AFE, .. AE=AF. And GBE=ABF=72° =BGE, .: GE=BE. Consider the lengths in the figure. Let AC=HE=1. Then, by above, AE=AF=15+1. And CE=AH=5, :: GE=BE=CE-CB= 15–1, :: CF2=AF2 AC2=5+2.75 and GH2=GE?H E2=5 – 2/5, .. EF2=CF2+CE2=10+2/5 and AG-=GH2+ AH2=10 – 2.15. Lastly,

FG=2. FO=2. AB=4,
.:: from A GFE, sin 18°=1(15-1), cos 18°=1(10+2/5),

from A FG A, sin 54°=$(5+1), cos 54°=1/(10– 2,75),
from A GAH, tan 18°

7(
5-2,75)

V(10 – 2,75)
sec 18°=
15

15
(5+ 22/5)

(10+2/5) from A FEC, tan 54° =

sec 54°= 15

15 from A AFC, cot 18°= [(5+2.75), cox 18°=15+1,

from A EGH, cot 54°=(5 – 2,75), cox 54°=J5 – 1. It

may be observed that the lengths below A E have the same form as the corresponding lengths above A E, but a minus sign instead of a plus sign. Thus the corresponding ratios of 18° and of 54o have the same algebraical form, except for the sign of the second term. More

sign occurs in the primary ratios of 18° and in the secondary ratios of 54°. This last fact may be connected with the results of Art. 114.

Example. Hence show that
cot 9°=/(5 +2./5)+5+1, and cot 27° =(5 – 2./5)+75 - 1.

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113. To find the limiting values of the ratios as the triangle of reference collapses into a straight line. The triangle of reference OMP collapses into a straight line,

when either the base OM or the
perp. MP coincides with the hyp.
OP, while the remaining side
vanishes. Thus,
as MOP decreases to 0°,
OM becomes equal to OP, and MP
vanishes;
as MOP increases to 90°,
MP becomes equal to OP, and OM
vanishes.

M

f Now the ratio of two finite equal lengths is unity :

f

0 The ratio of zero to any finite length is zero :

f 0

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0.

The ratio of any finite length to zero is infinity: 6 =

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[For a fraction varies in the same direction as its numerator, but in the opposite direction to its denominator. Hence, if the numerator decreases indefinitely, the fraction decreases indefinitely :-01f=0. But if the denominator decreases indefinitely, the fraction increases indefinitely –f/0=0.] Hence the limiting values of the ratios are

0, 1 or oc.
0
0

of
sin 0 -0: tan 0 0:

1:

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= oc: cox 0)

cos 0 =

cot 0
f
sin 90°

f
=1: tan 90°

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cos 90°

0
0

f
=0: cot 90°=-=0: cox 90°: 1.

f
In this table the student should observe:

(1) That, the angles 0 and 90° being complementary, any ratio of the one= the co-ratio of the other.

(2) Since, for certain limiting angles, the sin, cos, sec or cox equals 1, strictly we must say of sin and cos—not that they are always less than 1—but that they are never greater than 1: and of sec and coxnot that they are always greater than 1–but that they are never less than 1.

(3) The primary ratios have their minimum values at 0, and their maximum values at 90°.

The secondary ratios have their maximum values at 0, and their minimum values at 90°.

P

114. General changes in value of the ratios, as the angle changes from 0 to 90°. Let a line of constant length

z OP revolve from any position 01. It will describe a continuously increasing angle, as its extremity P traces out the quadrant of a circle.

Drop a perpendicular PM

upon 01.

Then, as the angle IOP increases from 0 to 90°, OP remains constant: MPO

M

M I increases : OM decreases. Now, the perp. MP is in the numerator of sin and tan;

but in the denominator of cox and cot. And the base OM is in the numerator of cos and cot;

but in the denominator of sec and tan.

Thus as the angle increases from 0 to 90°, the primary ratios, sin, tan, sec increase : while the secondary ratios, cos, cot, cox decrease.

From 0 to 90°, sin increases from 0 to 1: but cos decreases from 1 to 0. tan increases from 0 to oc: but cot decreases from oc to 0. sec increases from 1 to o: but cox decreases from oc to 1.

The results of some of the preceding articles exemplify these statements : e.g.

Ni
12
13

14 sin 0o = : sin 30o = sin 45°

sin 60o =

: sin 90°
2
2

2

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:

$ 3.

THE SOLUTION OF TRIGONOMETRICAL EQUATIONS.

A Trigonometrical equation is an equation connecting the ratios of some unknown angle which has to be found.

The solution generally requires the following three steps:

(1) Express the different ratios which occur in terms of one single ratio ; by means of the formulæ of this or the preceding chapter.

(2) Find the value of this one ratio by the purely algebraic process of solving an equation,

(3) Put down the acute angle, whose ratio has the value found, by applying inversely the results of this chapter on the values of the ratios of specific angles.

Example 1. Solve 2 sin 0=1.
(1) Here only one ratio occurs.
(2) The algebraical process gives sin 0=ž.
(3) Reference to Art. 108 shows that =30°.
Example 2. Solve 13 (tan 0+cot 6)=4.
1

1 (1) Since cotA= .. 13( tan 0+

- 4. tan a

tan A (2) .: 73. tan20+13=4 tan ,

4
... tan20 tan A 1,

13
4

2
2
4

1
.. tan-A -
tan 0 +

1
13

13 3 3'
2
... tan 6 -

+
13 13
3
1

1
.. tan A=

or 13 or 13

13 (3) Reference to Art. 108 shows that 0=60° or 30°.

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13

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