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Sin AOP2

P,N,

ON

OP2

P,N2, &c.

Let OP, OP, OP3, OP, represent the position of the revolving line at any period of revolution in the several quadrants respectively,

And let P, N, P,Ñ,, P.Ng, P.N. be the respective perpendiculars from the end of the revolving line upon the initial line.

Then P, N, P,N, P,N3, PLN, are respectively the perpendiculars corresponding to the angles generated.

Also, ON, ON, ON, ON, are respectively the bases of the right-angled triangles with respect to the angles in question. We have then in the second quadrant

2, cos AOP2

OP, tan AOP,

ON, It is therefore evident that the relations between the trigoñometrical ratios, which were proved to exist in Art. 7, also hold for angles in the second quadrant—that is, angles between 90° and 180°.

And in the same way we may show that they hold for angles in the third, fourth, or any quadrant.

And again, if we suppose the line to revolve in a negative direction, and take the position OP', we shall have PN the perpendicular corresponding to the negative angle AOP', and ON' the base.

P'N'

ON
Hence, sin AOP' = cos AOP!

OP
P'N'
tan AOP

&c.

ON And the relations proved in Art. 7 may be also similarly proved to exist here.

Hence the relations proved in Art. 7 hold for any angles whatever. Changes of Magnitude and Sign of the Trigonometrical

Ratios. 11. Let OP, OP, OP, OP, be positions of the revolving line in the several quadrants respectively; P N, P,N,

OP

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PN,

, cos AOP,

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0° ;

1

1,

1

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P.N3, P.N., the respective perpendiculars; and ON, ON, ON 3, ON the bases of the corresponding right-angled triangles.

Then,
(1.) In the first quadrant-

ON
Sin AOP
OP,

OPY
tan AOP, =

PAN,

&c.

OP At the commencement of the motion of the revolving line, the angle AOP

Also, the perpendicular P,N, = 0,
And the base ON

OP,
Hence, we have
0

OP * Sin 0°

0, cos 0°
OP

OP,
0
tan 0°

0.

OP, As the revolving line moves from OA towards OB, P,N, increases and ON, diminishes; and when it arrives at OB, we have PAN,

OP ,and ON = 0. But the angle generated is now a right angle. Hence we have

0
Sin 90°
1, cos 90°

1,
OP

OP, tan 90°

0 Hence, as the angle increases from 0° to 90°The sine changes in magnitude from 0 to 1 and is +. The cosine changes in magnitude from 1 to 0 and is +. The tangent changes in magnitude from 0 to co and is +.

(2.) In the second quadrantHere the perpendicular P,N, is +,

and the base ON, is * The student ought properly to look upon the values 0, 1, 0 here obtained as the limiting values of the sine, cosine, and tangent respectively, when the angle is indefinitely diminished.

OP

1

1

OP1

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2

2

Hence the sine during the second quadrant is +, the cosine is –, and the tangent is

Again, as the revolving line moves from OB to OA', the perpendicular P,N, diminishes until it becomes zero. Also, the base ON, increases in magnitude, until it finally coincides with OA', and .. OP, But the angle now described is 180°. Hence we have 0

OP,
Sin 180°

0, cos 180°
OP

OP

0 tan 180°

0, &c.

2

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2

2

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OP,

Hence in the second quadrantThe sine changes in magnitude from 1 to 0, and is positive. The cosine changes in magnitude from 0 to 1, and is negative. The tangent changes in magnitude from a to 0, and is negative.

And in the same way may we trace the changes of magnitude and sign in the third and fourth quadrants.

Thus we shall find

(3.). In the third quadrantThe sine changes in magnitude from 0 to 1, and is negative. The cosine changes in magnitude from 1 to 0, and is negative. The tangent changes in magnitude from 0 to co, and is positive.

(4.) In the fourth quadrantThe sine changes in magnitude from 1 to 0, and is negative. The cosine changes in magnitude from 0 to 1, and is positive. The tangent changes in magnitude from 0 to 0, and is negative.

Moreover, as the cosecant, secant, and cotangent are respectively the reciprocals of the sine, cosine, and tangent, it follows that their signs will follow respectively the latter, and that their magnitudes will be their reciprocals.

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CHAPTER IV. TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES

OF THE TRIGONOMETRICAL RATIOS OF 30°, 45°, 60°, &c. 12. To prove that sin A = cos (90° A), and that

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sin (90° A).

cos A

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sec A cosec A

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Using the same figure as in Art. 5, we have

PM Sin A

= cos APM.

AP But LAPM 90° A, ... Sin A = cos (90° - A), and similarly

sin (90° – A), tan A cot (90°

A),
cot A
tan (90°

A),
cosec (90° – A), A

-B

M = sec (90° - A). 13. Ratios of 45°.

In the last figure, suppose Z PAM 45°, then also LAPM 90° 45° 45°. And hence Z PAM = L APM, and ... PM AM (Euc. I., 6). Hence, also— AP or VAM” + PM N2

AM’ or V2 PM”. .. AP = AM J2 or PM 2. Hence we have

PM PM 1 Sin 45° sin PAM

cos 45°, AP

by PM 12 2 Art. 8.

PM PM Tan 45o = tan PAM

1 cot 45°, by Art. 8. AM PM

AP AM 12 Sec 45° = sec PAM

12 AM

= cosec 45°,

AM by Art. 8. 14. Ratios of 30° and 60°. In the same diagram, suppose Z PAM

30', then 2 АРМ 90° - 30° = 60°.

Hence, if we conceive another triangle equal in every respect to APM to be described on the other side of AM, the whole would form an equilateral triangle whose side is AP.

Hence, PM = { AP.

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AP

=

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Р
1

We hence have

PM Sin 30o = sin PAM

AP

= cos 60°, by Art. 8. AP

13 AM

AP Cos 30°

2 = cos PAM

13

= sin 60°, by
AP AP
Art. 8.
PM
AP

1 Tan 30°

* V3 = cot 60°, by

' AM 13

-AP

2 Art. 8. 15. To show that sin (180o – A) = sin A,

°

A
cos (180° - A) cos A,

Let LAOP A,
P.

P.

And let the revolving line describe an

angle AOP, = 180° A' N2

- A;

Then LAOP, = 180° - (180° - A) = A;

Hence, 2 AOP, = A OPz.

Hence, also (Euc. I., 26), if P,N, P,N, be drawn perpendicular to AA', P N = P2N2, ON2 ON,

We have therefore-
Sin (180° - A) = sin AOP, =

sin A.
ОР, OP,

ON, Cos (180° – A)

cos AOP, And similarlyTan (180° - A) tan A, cot (180° AS - cot A. Sec (180° - A) = sec A, cosec (180° - A) = cosec A.

2

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1

1

1

P.N2 =

PAN,

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ON
ОР,

2

OP

COS A.

1

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