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If we are given the value of the sine of an angle, it is important to be able to find all the angles which have that value for their sine.

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38. General Expression for All Angles which have a Given Sine a. Let O be the centre of the unit circle. Draw the diameters AA', BB', at right angles. From O draw on OB a line ON, so P2

IN that its measure is a. Through N draw PP' parallel to M

M)
AA'. Join OP, OP', and draw PM,
P'M', perpendicular to AA'.
Then since MP

ON
A,

B' the sine of AOP is equal to the sine of AOP'.

Hence the angles AOP and AOP' are supplemental (Art. 30), and if AOP be denoted by a, AOP' will = — a.

Now it is clear from the figure that the only positive angles which have the sine equal to a are « and a the angles formed by adding any multiple of four right angles to a and a Hence, if o be the general value of the required angle, we have 0= 2n1 + Q, or 0= 2 107 + - (l,

(1) where n is zero or any positive integer.

Also the only negative angles which have the sine equal to a are – (+), and (27 — «), and the angles formed by adding to these any multiple of four right angles taken negatively; that is, we have

0=217 - (+«), 0= 2NT – (21 – a), (2) where n is zero or any negative integer.

Now the angles in (1) and (2) may be arranged thus: 2n7 + a, (2n +1) - O, (2n – 1). — a, (2n — 2) + a, +

* A all of which, and no others, are included in the formula 0=NT +(-1)",

(3)

- O.

=

T

T

T

where n is zero, or any positive or negative integer. Therefore (3) is the general expression for all angles which have a given sine.

NOTE. — The same formula determines all the angles which have the same cosecant as a.

a

A!

M

39. An Expression for All Angles with a Given Cosine a.Let O be the centre of the unit circle.

B
Draw AA', BB', at right angles.

Р
From O draw OM, so that its meas-
ure is a.
Through M draw PP parallel to

А.

|М BB'. Join OP, OP'.

Then since OM = a, the cosine of AOP is equal to the cosine of AOP'.

B' Hence, if AOP

=, AOP= Now it is clear that the only angles which have the cosine equal to a are a and - 0, and the angles which differ from either by a multiple of four right angles.

Hence if & be the general value of all angles whose cosine is a, we have

0= 2nd † a,

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=

where n is zero, or any positive or negative integer.

NOTE. — The same formula determines all the angles which have the same secant or the same versed sine as a.

40. An Expression for All Angles with a Given Tangent a. - Let O be the centre of the unit circle. Draw AT, touching the circle at A, and take AT so that its measure is a. Join OT, cutting the circle at PA'

А. and P'. Then it is clear from the figure that

P' the only angles which have the tangent equal to a are a and + + a, and the angles which differ

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from either by a multiple of four right angles. Hence if a be the general value of the required angle, we have 0 = = 2n1 +0, and 2 nu + 7 +0.

(1) Also, the only negative angles which have the tangent equal to a are – (1 — «), and —(21 «), and the angles which differ from either by a multiple of four right angles taken negatively; that is, we have

0=217 -(1-a), and 2n7 - (27 –), . . (2) where n is zero or any negative integer.

Now the angles in (1) and (2) may be arranged thus:

2 17 + 0, (2n +1) + + a, (211 – 1) + +, (2n — 2) + + a, all of which, and no others, are included in the formula

=

7T

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0 = na to,

(3) where n is zero, or any positive or negative integer. Therefore (3) is the general expression for all angles which have a given tangent.

NOTE. - The same formula determines all the angles which have the same cotangent as a.

EXAMPLES.

1. Find six angles between – 4 right angles and + 8 right angles which satisfy the equation sin A = sin 18°.

We have from (3) of Art. 38,

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0=na +(-1)", or A=n x 180°+(-1)"18°.

10'

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Put for n the values -2, -1, 0, 1, 2, 3, successively, and we get A =– 360° +18°, – 180° - 18°, 18°, 180° - 18°, 360° + 18°, 540° — 18°;

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that is,

- 342, - 1989, 189, 162°, 378°, 522o.

NOTE. — The student should draw a figure in the above example, and in each example of this kind which he works.

equations (1) sin A = (2) sin A =

V3

2. Find the four smallest angles which satisfy the

1

1 )

(3) sin A V2

2 1 (4) sin A

2
Ans. (1) 30°, 150°, -210°, - 330°;

°

(2) 45°, 135°, – 2250, – 315o;

(3) 60°, 120°, - 240°, – 300°;

3 (4) -30°, – 150°, 210°, 330°.

41. Trigonometric Identities. - A trigonometric identity

is an expression which states in the form of an equation a relation which is true for all values of the angle involved. Thus, the relations of Arts. 13 and 23, and all others that may be deduced from them by the aid of the ordinary formulæ of Algebra, are universally true, and are therefore called identities, but such relations as sin @= }, cos (= },

0 are not identities.

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2. Prove that cot 0 – sec ô cosec 0 (1 – 2 sin’6)= tan 0. cot 0 - sec 0 cosec 0 (1 – 2 sinề0)

1 . 1

(1 – 2 sin’0) (Art. 24) sin cos A sin 0

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cos0 1+2 sin? 0

sin A cos 0

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NOTE. - It will be observed that in solving these examples we first express the other functions in terms of the sine and cosine, and in most cases the beginner will find this the simplest course. It is generally advisable to begin with the most complicated side and work towards the other.

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5. (tan 0 + cot ) sin 0 cos 0 = 1.
6. (tan 0 — cot 6) sin 0 cos 0 = sino 0 – cos? 0.
7. sinA : cosec-0 = sin.
8. sec 0 — tant0 = seco0 + tan20.
9. (sin 0 - cos 6)=1-2 sin cos e.

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1 + cos 0 11.

(cosec 0 + cot 6)? 1- cos 0 12. (sin 0 + cos 0)2 + (sin 0 — cos 6)?= 2. 13. sin4A - costo = sino - cos”0.

2

14. sino + vers = 2 (1 – cos 6).

15. cot? 0 – cos'0 = cot’O cos20.

EXAMPLES.

In a right triangle ABC (see figure of Art. 15) given: 1. a=p? +Pq, c= q*+Pq; calculate cot A.

.

P

Ans. Vop.

n2

2. b=lm = n, c= ln = m; calculate cosec A.

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