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53. If the Sine of an Angle be given, the Sine and the Cosine of its Half are each Four-Valued.

2sines

We have

X
COS

(Art. 49)

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and

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(Art. 23)

2

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+ cos

=1+ sin x.

2

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COS

1– sin a.

2

* = sin æ

+ cos” = 1 By addition, (sin ) By subtraction, (sin

:. sin + cos= V1+ sin æ

sin - cos = + V1 – sin 2 .. 2 sin" = V1+sin x + V1 – sin æ ::

os

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2

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( 3 (4)

(3)

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2

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Thus, if we are given the value of sinx (nothing else being known about the angle x), it follows from (3) and

X ) (4) that sin, and cos have each four values equal, two

2

2 by two, in absolute value, but of contrary signs.

The case may be investigated geometrically thus:

Let ON = the given sine (radius being unity) = sinx. Through N draw PQ parallel to OA;

B and draw OP, OQ. Then all angles whose sines are equal to sin x are terminated either by OP or OQ, and

A the halves of these angles are terminated by the dotted lines Op, Oq, Or, or Os. The sines of angles ending at Op, Oq, Or, and Os are all different

B'

Р

A

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2

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2

in value; and so are their cosines. Hence we obtain four values for sin, and four also for cos

in terms of x. 2

2 When the angle x is given, there is no ambiguity in the calculations ; for is then known, and therefore the signs and relative magnitudes of sin" and cos, are known. Then equations (1) and (2), which should always be used, immediately determine the signs to be taken in equations (3) and (4) Thus, when lies between – 45° and + 45°, cos

** >

> sin and is positive.

Therefore (1) is positive, and (2) is negative · and hence (3) and (4) become

2 sin

m = V1+ sin æ – V1 – sin æ, 2cos. = V1+sinx+ V1 – sin æ.

2 When lies between 45° and 135°, sin >cos, and is

° °, 2 positive.

Therefore (1) and (2) are both positive; and hence (3) and (4) become

2 sin = V1 + sinæ + V1 – sinæ,

2

=

2

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2 cos

cose

V1 + sinx – Vĩ– sin a.

2

And so on.

54. If the Tangent of an Angle be given, the Tangent of its Half is Two-Valued.

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tang

Thus, given tan 6, we find two unequal values for tan

2 one positive and one negative.

This result may be proved geometrically, an exercise which we leave for the student.

55. If the Sine of an Angle be given, the Sine of OneThird of the Angle is Three-Valued.

We have sin 3x = 3sin x - 4sinox (Art. 50)

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a cubic equation, which therefore has three roots.

EXAMPLES.

1. Determine the limits between which A must lie to satisfy the equation

2 sin A=-V1+sin 2 A – V1 – sin 2 A.

By (1) and (2) of Art. 53, 2 sin A can have this value only when

sin A +cos A - V1 + sin 2 A,

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Therefore A lies between 225o and 315°, or between the angles formed by adding or subtracting any multiple of four right angles to each of these; i.e., A lies between

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where n is zero or any positive or negative integer.

2. Determine the limits between which A must lie to satisfy the equation

2 cos A = V1 + sin 2 A - V1 – sin 2 A.

By (1) and (2) of Art. 53, 2 cos A can have this value only when

cos A + sinA= V1+ sin 2 A, and cos A – sin A =- V1 – sin 2 A;

i.e., when sin A >cos A and positive.

Therefore A lies between

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where n is any positive or negative integer.

3. State the signs of (sin 0 + cos ) and (sin 0 - cos ) when @ has the following values: (1) 22°; (2) 191°; (3) 290°; (4) 345°; (5) – 22°; (6) – 275°; (7) – 470°;

; 5)

(8) 1000

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2

Ans. (1) +, -; (2)
-, +; (3) -; (4) +,

- ;
(5) +, -; (6) +, +; (7) -,-; (8) -, -.

6 4. Prove that the formulæ which give the values of sin and of cos in terms of sin x are unaltered when a

2
has the values

(1) 92°, 268°, 900°, 4n+ + , or (4n+2)+ - *;
(2) 88°, – 88°, 770°, – 770°, or 4nn †
,

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5. Find the limits between which A must lie when

2 sin A = V1 + sin 2 A - V1 - sin 2 A.

56. Find the Values of the Functions of 223 - In (3),

(4), and (5) of Art. 51, put x = 45°. Then

1

V2 - V2 sin 221°=

2

2

cos 45°

=

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Since 22° is an acute angle, its functions are all positive.

The above results are also the cosine, sine, and cotangent respectively of 673", since the latter is the complement of 221° (Art. 15).

57. Find the Sine and Cosine of 18°.

18°; then 2x = 36°, and 3x = 54°.

Let x =

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Solving the quadratic, and taking the upper sigu, since sin 18° must be positive, we get

15-1 sin 18° =

4

=

Also, cos 18°= V1 – sin’18°

V10 +2 Võ

4 Hence we have also the sine and cosine of 72° (Art. 15).

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