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Let XOE be the angle A, and EOF the angle B, then XOF is the angle (A + B).

In OF, the line which is one of the sides of the angle (A+B), take any point P, and from P draw PM and PN at right angles to OX and OE respectively. Draw NH and NK at right angles to MP and OX respectively. Then the angle

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1 Or thus. On OP as diameter, describe a circle. This will pass through M and N, because the angles OMP and ONP are right angles; therefore MPN and MON are angles in the same segment; so that the angle MPN=MON=A.

Let XOE be the angle 4, and FOE the angle B. Then in the figure, XOF is the angle (A – B).

In OF, the line which bounds the compound angle (A — B), take any point P, and from P draw PM, PN at right angles to OX and OE respectively. Draw NH, NK at right angles to MP and Ox respectively. Then the angle

NPH = 90° – HNP = HNE = XOE = A‚1

Now sin (A- B)= sin XOF =

=

=

KNON

ON OP

= sin XOE⋅ cos FOE

=

= sin A cos Bcos A sin B.

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PH · NP

NP. OP

1

Also, cos (4 – B) = cos XOF = OM

OP

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MP
OP

2. Show that sin 15°

EXAMPLE. Find the value of sin 75°.

sin 75° = sin (45° + 30°)

1. Show that cos 75° =

=

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OK. ON NH.NP OK ON NH NP
+
+
ON OP NP OP ON OP NP OP

=

KN ON PH_NP
ON OP NP OP

cos HPN sin FOE

= cos XOE cos FOE+ sin HPN sin FOE

= cos 4 cos B + sin 1 sin B.

.

√3-1

2√2

√3-1
2√2

MH - PH KN
OP

OP

=

sin 45° cos 30° + cos 45°. sin 30°

1

OK + KM
OP

1 V3 1

+

V2

√2

√3+1 _ √2 (√3 + 1).

=

EXAMPLES. XVII.

OK NH = + OP OP

PH

OP

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1 Or thus. On OP as diameter, describe a circle. This will pass through M and V, because the angles OMP and ONP are right angles; therefore the angles MPN and MON together make up two right angles, so that the angle HPN = MON = A.

5. If sin A = and sin B = }, find a value for sin (A+B) and for cos (AB).

6. If sin A = .6 and sin B = 3, find a value for sin (A+B) and for cos (A + B).

7. When sin A =

or

1

M

and sin B

H

8. Prove that sin 75° = .9659 ...

9. Prove that sin 15° = .2588 ..

10. Prove that tan 15° = .2679...

11. Calculate sin 90° and cos 90°, using the ratios of 45°.

*78. The proofs given in Arts. 76 and 77 are really, at least so far as the figures are concerned, rigorous only in case (A + B)< 90°, and hence A and B are each less than 90°. By a careful regard as to sign of angles and lines, however, the wording will hold for angles of any magnitude. The student may satisfy himself that this is true by constructing suitable figures. The accompanying figure will serve in case A+B < 180° and A< 90°, B> 90°.

K

1

V10

N

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B

then one value of (A + B) is 45°.

FIG. 48.

A

*79. In order, however, to remove the restrictions as to magnitude, placed on A and B, we shall pursue the following course:

Suppose

180° > A > 90° and B < 90°.

Put

A = 90° + A'; then A'<90°.

=

sin (A+B) sin (90° + A' + B)= sin [90° +(A' + B)], (Art. 64.) sin [90° + (A' + B')]= cos (A' + B). (Art. 62.)

A' and B are each less than 90°. We may therefore write,

cos (4' + B)= cos A' cos B- sin A' sin B; (Art. 76.)

cos (A — 90° + B)

= cos(4-90°) cos B - sin (A - 90°) sin B;

since

cos (A — 90° + B) = cos [90° − (A + B)]

-

= sin (A+B);

sin (4 — 90°) — — sin (90° — A)

--

cos A,

sin (A + B)= sin A cos B+ cos A sin B.

By a repetition of this process A may become an angle of any quadrant, and it is evident the reasoning and wording remain just as above; so that the magnitude of A is unrestricted.

.. sin(A+B)= sin A cos B + cos A sin B,

whatever be the magnitude of A, provided B< 90°. There remains yet to inquire what happens if A and B are both greater than 90°.

Suppose

and since

we may write,

Put

180° > B > 90°.

B=90° + B'; then B'< 90°;

sin (A+B) sin (90° + A+ B')= cos (A + B').

=

(Art. 58 Cor.)

(Art. 61.)

(Art. 58.)

(Art. 61.)

We can now apply (i.), whence

sin (A+B)= cos A cos B' - sin A sin B'
= cos A sin B+ sin A cos B.

(i.)

Equation (1) is therefore true if B> 90° but <180°. As before we need only repeat the process in order to remove all restrictions as to the magnitude of B. Hence we conclude that whatever the values of A and B,

sin (A+B) = sin A cos B + sin B cos A.

By a similar process, the three kindred formulæ may be shown to be true for all angles.

NOTE. See Nixon's Elementary Trigonometry, Sec. 18.

For an elegant proof of this theorem by the method of projections, see Hobson's Plane Trigonometry, Sec. 40.

80. It is important that the student should become thoroughly familiar with the formulæ of Art. 75, and that he should be able to work examples involving their use.

Prove the following statements:

6. tan atan 8 =

Dina

Ga

7. tan a

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1. sin (A + B) + sin (A –
2. sin (A + B) – sin (A –
3. cos (A + B) + cos (A –
4. cos (AB) — cos (A + B) = 2 sin A. sin B.

B) = 2 sin A⋅ cos B.
B) = 2 cos A. sin B.
B) = 2 cos A⋅ cos B.

-

5.

=tan A.

sin (A+B) + sin (A − B)
cos (1+B) + cos (A − B)
sin (a + B).
COS a. cos B

13.

tan 8 =

tan B

10. tan a cot ß =

tan

tan

tan 0 tan

tan

1 tan

20.

=

sin (a - B).
cos acos B

23.

cos (a - B).
sin a cos B

·

tan +1
tan

EXAMPLES. XVIII.

cos (a + B)
sin a cos B

cos (a - B)
cos asin B

1

sin (@ +

sin (@

=

cos (0-4).

cos (0 + $)
cot &
cot Y
coty cot 8 + 1

=

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).

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tan2 8 =

14.

=

15.

16.

17.

18.

19.

tan @+cot.
cot-tan @

cot + coto sin (@+).
cot o
sin (0-0)

cot Ꮎ

tan (y - 8).

tan cot +1
1
tan . cot
1+coty tan d
cot Y tan d

1- coty tan d
cot ytan d
tan cot d
Y
1
tany+cot &
tany cot 8 +1
cot &
tan y

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=cos(0-4). sec (0+).

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sin (a + B) sin (a – B).
cos2 a.
cos2 B
cos (a + B) cos (a
sin2 a⚫
· cos2 B

=tan (a + B) · tan (a — B).

22. cot2 a

1

tan2 a
tan2 B
tan2 a tan2 B
24. sin (a + B)
25. cos (a + B)

· sin (a — 8) = sin2 a
⋅ cos (a — B) = cos2 a
sin A-
√2
27. √2. sin (A + 45°) = sin A + cos A.
28. cos Asin A = √2 ⋅ cos (A + 45°).

cos A

26. sin (A-45°)

29. cos (A+45°) + sin (A — 45°) = 0.

30. cos (445°) sin (A + 45°).

31. sin (+) • cos 0 ·cos (+)· sin = sin .
32. sin (6) $) · cos + cos (0 − p) · sin ☀ = sin 0.
33. cos (+)· cos + sin (0 + ø) · sin 0 = cos p.

tan (0) + tan

34.

=tan 0.

tan (0-4). tan

-

B).

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— sin2 ß = cos2 ß

cos2 a.

sin2 B cos2 B - sin2 a.

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