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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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B

In the figure of Art. 120, draw CA' at right angles to CA. Then LNCA' = comp.

of LNCA =NAC = A, and int. 2 BC A' = comp. of ext. 2 C = comp. of (A + B).

2A ABC NC.BA
.. cos (A + B) = sin BC A'

A'C.BCA'C. BC
NC.(BN A'N) NC BN A'N NC

AC.BC A'C' BC ACBC = 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 ENCA
= comp. of _NCA =

- NAC = A, and

BCM = comp. of int. 2 C = comp. of (A B).

ext. 2

2 - ABC NC. BA
:. cos (A - B) = sin BCM =

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

AC.BC AC BC AC BC
= cos A. cos B + sin A. sin B.

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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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Let POC=A; 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 + A COQ = 2 - POR and POC ACOQ = 2 - ROC. _ POC + + COQ=2 _ POR and 2 POC COQ = 2 _ ROC. 2 A POC 2 A COQ

4 Δ POR .. sin A + sin B

OP.OC" OQ.OC OP.OC
2PR OR
OP 'oc
2 sin ! (A + B) cos (4-B),

2 A POC 2 ACOQ 4 A ROC and sin A - sin B=

OP.OCOQ .OC

09.0c OP.ос
20R RC
OP'oc
= 2 cos } (A + B) sin } (A B).

=

125.

To prove the formula
cos B +cos A = 2 cos 3 (A + B) cos } (A B)....... (7),
cos B-cos A = 2 sin } (A + B) sin 3 (A B)....... (8).

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In the figure of last article, draw FOC' at right angles to OC.
Then 4
OC'R= = comp.

of 2 COR _ ROC = 1 (A B), and COQ=comp. of - QOF, and _ POC = comp. of C'OP.

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

+

2 A POC" OP. OC'

2 cos 1 (A + B) cos Ž (A B).

2 A QOC" And cos B -cos A = sin QOF - sin COP =

OQ.OC"
POR 2PR RO

OQ. OC" OPOC
= 2 sin } (A + B) sin } (A - B).

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,
:: LRPM = complement of QPV = _ RQP=A.
Thus

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

2 sin A.cos A. OP QR QP QR

201 QM- MR QM QP MR RP (10) 20P QR

QP' QR RP' QR

= cos” A - sino A. OP QR QM + MR cot A + tan A (11) cosec 2A MP2P 2MP

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

2
MR PR MR OR - OM 1
(13) sin? A
PR' QR QR 20P

2
QP Q0 + Ом 1 + cos 2 A
(14) Cos’ A
QP QR

20P

2
MR OR-OV ОРОМ
(15) tan A
MP

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

QM QO+OM ОР OM (16) cot A

=cosec 2A +cot2A. MP MP

MP MP

=

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QR

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Similarly we may show that
2 tan A

seco A
(17) tan 2A

(18) sec 2A 1-tanA

2-secA' (19) cos 2 A

2 cos? A -1. (20) cos 2A = 1 – 2 sino A. 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 :

..(22),

tan (A B) = 1+tan A tan B

+

cot A cot B-1

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

.(23),

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

cot B – cot A

(24).

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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. 129–134.

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 A = 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).

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