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48. To find the values of the functions of 180° + A in terms of those of A.

By Arts. 45 and 44, we have:

sin (180°+4)= cos (90°+ A) = — sin A.

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49. To find the values of the functions of 270° — A in terms

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50. To find the values of the functions of 270° + A in terms

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tan (270°+4)=-cot (180°+4)= cot A.

A)

cot (270°+4)=-tan (180°+4)=-tan A.

sec (270°+A) = — csc (180°+A) =

csc (270°+4)=

=

csc A.

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51. To find the values of the functions of 360°- A and 360°+ A in terms of those of A.

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By Art. 34, the functions of 360°- A are the same as those of - A, and the functions of 360°+ A are the same as those of A.

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52. The operation illustrated in Art. 51 may be extended indefinitely; we should find for the angles 450°- A, 810°— A, -270° - A, etc., the same results as for 90°-A; and so on.

53. The results of Arts. 42 to 52 may be expressed in the following rule, which is derived by inspection from the formulæ of Arts. 42 to 51:

Any function of 0°, or an EVEN multiple of 90°, plus or minus A, is the SAME function of A; and any function of an ODD multiple of 90°, plus or minus A, is the COMPLEMENTARY function of A.

To obtain the algebraic sign, regard A as an acute angle, and apply the table of Art. 37.

1. Required the value of sec (990°+ A).

Since 990° is an odd multiple of 90°, the absolute value of the result is csc A.

And if A is an acute angle, 990°+A is in the fourth quadrant, in which the sign of the secant is positive.

Hence,

sec (990°+4)= csc A.

2. Required the value of tan ( — 180° — A).

Since 180° is an even multiple of 90°, the absolute value of the result is tan A.

And if A is an acute angle, — 180°- A is in the second quadrant, in which the sign of the tangent is negative.

Hence,

tan (— 180° — A) — — tan A.

EXAMPLES.

Find the values of the functions of the following angles in terms of the functions of A:

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54. The functions of any positive or negative angle whatever may be expressed in terms of the functions of an angle between 0° and 90°.

1. Express sin 317° as a function of an angle between 0° and 90°.

or

We have by the rule of Art. 53,

sin 317° sin (270° + 47°) = cos 47°,

=

==

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Express the following in terms of the functions of angles between 0° and 90°:

2. cos 152°.

3. tan 522°.

4. sec (77°).

5. csc 230°.

6. cot (-129°). 7. sin 865°.

Express the following in terms of the functions of angles

between 0° and 45° :

8. cot 83°.

10. sec 165°.

12. tan 520°.

9. sin (-50°). 11. cos (-303°). 13. csc 768°.

55. The following table gives, for convenient reference, the functions of the angles 0°, 30°, 45°, 60°, 90°, 120°, etc.

The values of the functions of 0°, 30°, 45°, 60°, 90°, 180°, 270°, and 360° have already been proved in Arts. 16, 17, 38, 39, 40, and 41; the others may be derived by aid of Art. 53, and are left as exercises for the student.

As an illustration, we will give the proof for cot 150°.
By the rule of Art. 53:

cot 150° cot (180°-30°) cot 30° -√3 (Art. 17).

=

=

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56. Given the value of one of the functions of an angle, to find the values of the remaining functions. (Compare Art. 15.)

=

1. Given sin A ing functions of A.

3

; required the values of the remain

The example may be solved by a method similar to that employed in Art. 15.

Since the sine is the ratio of the ordinate to the distance, we may regard the point of reference as having its ordinate equal to -3, and its distance equal to 5.

There are two such points, P and P', and consequently two angles, XOP and XOP', in the third and fourth quadrants respectively, either of which satisfies the given condition.

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Then OM = OM' = √ÕP2 – PM2 = √25

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co-ordinates of P are (— 4, −3), and of P', (4, − 3).

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