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122.

To prove the formula

cos (A + B) = cos A cos B – sin A sin B.... ...... (3).

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In the figure of Art. 120, draw CA' at right angles to CA.

Then - NCA' = comp. of NCA = 1 NAC = A, and int. 2 BC A' = comp. of ext. 2 C = comp. of (A + B).

Α' BC NC.BA
.. cos (A + B) = sin BCA

A'C. BCA'C. BC
NC. (BN-A'N) NC ON AN NC

.
A'C. BC AC BC "AC BC

'
= cos A. cos B-sin A. sin B.

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123. To prove the formula

cos (A B) = cos A cos B + sin A sin B..... ... ... (4).

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In the figure of Art. 121, draw MCA' at right angles to CA.

Then NCA' = comp. of - NCA= _ NAC = A, and ext. - BCM = comp. of int. 20 comp. of (A B).

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2 A A'BC NC. BA'
1. cos (A – B) = sin BCM

AC.BC A'C.BC
NC (BN+NA') _ NC BN NA' NC

AC.BC A'C' BC AC · BC
= cos A. cos B + sin A. sin B.

+

.

124.

To prove the formula
sin A + sin B = 2 sin } (A + B) cos } (A - B).......(5),
sin A – sin B = 2 cos } (A + B) sin } (A - B)....... (6).

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A

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Let _ POC = 4; and QOC=B. Take OP=OQ. Join PQ, cutting OC in C.

Let OR bisect angle POQ, triangle POQ, and base PQ. Then (Art. 116, I.) A POC + ACOQ = 2 - POR and A POC COQ = 2 - ROC. _ POC + 2 COQ=2 _ POR and _ POC - + COQ = 2 _ ROC. 2 A POC 2 ACOQ 4 A POR

4Δ .. sin A + sin B

OP.OC
0.00

ОР. ОС
2PR OR
OP OC
- 2 sin } (A + B) cos } (A B),
2 A POC 2 ACOQ

4 A ROC and sin A - sin B =

OP.OCOQ.00
ОР

OP. OC
20R RC

OP 'OC
= 2 cos } (A + B) sin } (A - B).

.

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cos B + cos A = 2 cos } (A + B) cos } (A B)....... (7), cos B-cos A = 2 sin } (A + B) sin } (A – B)....... (8).

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In the figure of last article, draw FOC' at right angles to OC. Then _OC'R= comp. of _C'OR _ ROC

(A

- B), and COQ = comp. of 2 QOF, and _ POC = comp. of <C'OP.

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Also (Art. 116, II.), A QOC" + - POC" = 2 - ROC" and A QOC' A POC" = 2 POR,

2 A QOC" 2 A POC .. cos B+cos A = sin QOF+ sin C'OP=

OQ. OC" OP. OC"
4 A ROC 2RO C'R
OP. OC" OP '00

2 cos 1 (A + B) cos } (A B). And cos B - cos A = sin QOF - sin COP =

2 A QOC" 2 A POC"

OQ. OC' OP. OC'
POR 2PR RO
OQ. OC"

OP '00
= 2 sin } (A + B) sin } (A B).

A

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126. To find the relations between the ratios of an angle and those of its half or double.

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Describe a semicircle QPR, centre 0, on diameter QR.
Make - RQP = A : and draw PM perpendicular to QR.

Then < ROP at the centre = 2.2 RQP at circumference = 2A, and

LQPR in semi-circle = a right-angle, :. - RPM = complement of QPN = RQP= A. Thus

MP 2HP 2P QP
(9) sin 2A

- 2 sin A.cos A.
OP QR QP QR
201 QV - MR
QV - MR QU QPMR RP

QM
(10) cos 2A
20P QR

QP QR RP' QR

= cos? A - sino A. OP QR Q + VR cot A + tan A (11) cosec 2A MP 2MP 2MP

2 OM 201 QM JR cot A - tan A (12) cot 2A MP 2MP 2 MP

2 MR PR MR OR-O 1 (13) sino A PR' QR QR 20P

2 QvQPQM QO+OM 1 + cos 2 A (14) cosA QPQRQR 20P

2 MR OR-O OP ОМ (15) tan A

=cosec 2A-cot 2 A. MP MP MP MP

90 + OM OP ON (16) cot A

=cosec 2 A +cot 2A. MP IP UP MP

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cos 2 A

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.

Similarly we may show that
2 tan A

seco A (17) tan 2A

(18) sec 2A 1-tan? A

2 - seco A (19) cos 2 A = 2 cosa A -1. (20) cos 2A = 1 – 2 sinA.

127. To find the tangent and cotangent of the sum or difference of two angles.

sin (A + B) sin A cos B + cos A sin B
tan (A + B)
cos (A + B) cos A cos B - sin A sin B

[Arts. 120, 122. In order to express this fraction in terms of tan A and tan B, we must divide its numerator and denominator by cos A.cos B.

tan A + tan B .. tan (A + B)

(21). 1-tan A tan B

tan A – tan B Similarly : tan (A B) =

.(22), 1 + tan A tan B

cot A cot B - 1 cot (A + B) =

(23), cot B + cot A

cot A cot B + 1 cot (A B) =

(24). cot B - cot A

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128. The preceding 24 formulæ are not independent of one another. It will be useful to show their algebraical interdependence. This is done in Arts. 129–138.

. 129. Given sin (A + B) = sin A cos B + cos A sin B ..(1), to prove

cos (A – B) = cos A cos B + sin A sin B ...... (4). In (1) let A = 90° – C. Then

sin A = sin (90° — C) = cos C,
and

= cos (90° – C) = sin C,
and sin (A + B) = sin (90° – C B) = cos (C B).

:: (1) becomes cos (C B) = cos C cos B + sin C sin B.
Writing A for C we have (4). Similarly
Given sin (A – B) = sin A cos B - cos A sin B...........

(2), we can prove cos (A + B) = cos A cos B - sin A sin B

:(3).

COS A

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