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I. Then let KOF = A, then IOF= A + 90°. .. (1) becomes cos (A + 90°) = - sin A

(3), (2) becomes sin (A + 90°) = + cos A

(4). II. Let IOF-= A, then KOF = A - 90°. :: (1) becomes sin (A – 90°)=-cos A

(5), (2) becomes cos (4 – 90°) = + sin A...... (6). 327. These formulæ may be applied to find the ratios of A = 180°, A + 270° &c., in terms of ratios of A. Thus sin (A + 180°) i.e. sin{(A + 90°) + 90°} = cos (A +90°) by (4)

sin A by (3), cos (A + 180°) i.e. cos {(A + 90°) + 90°=- sin (A + 90°) by (3)

cos A by (4), and so on.

328. In the next articles we give the usual proofs of the formulæ for the sine and cosine of the sum and difference of two angles, confining ourselves to positive acute angles.

329. To find the sine, cosine, and tangent of the sum of two angles, in terms of the ratios of the angles.

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Let XOY= A, and YOZ = B, then XOZ= A + B.

In OZ, the final line of the compound angle A + B, take any point P.

From P, draw PM, PQ perpendiculars upon 0X, OYthe initial and final lines of A—respectively.

From (, draw QN, QR perpendiculars upon OX, MP respectively.

Then LQPR = 90° - LRQP= 2 OQR= _XOQ = A.

[Or: RP, QP being perpendiculars upon 0X, OY contain the same angle as OX, OY.]

MP MR + RP NQ RP (1) sin (A + B) = sin X0Z=

ОР OP OP'OP

ОР =[introducing required hyps. under the numerators]

NQ OQ RP QP

OQOPQP OP
= sin NOQ. cos QOP + cos QPR, sin QOP
= sin A . cos B + cos A. sin B.

OM ON - MN ON RQ (2) cos (A + B) = cos XOZ=

OP OP OP OP = [introducing required hyps. under the numerators]

ON OQRQ QP

OQOPQP' OP
= cos NOQ . cos QOP - sin QPR . sin QOP

= cos A . cos B - sin A. sin B.
Since the triangles NOQ, RPQ are similar,

RP QP
..
ON OQ

MP NQ + RP (3) tan (A + B) = tan XOZ=

OM ON-RQ
NQ

RP
ON ON tan A + tan B
ON RQ
RQ RP

1-tan A. tan B ON RP ON

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- tan B,

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330. To find the sine, cosine, and tangent of the difference of two angles in terms of the ratios of the angles.

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Let XOY = A, and ZO Y = B. Then XOZ= A - B.
In OZ, the final line of the compound angle A B, take any

point P.

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From P draw PM, PQ perpendiculars upon 0X, OY-the initial and final lines of A—respectively.

From Q draw QN, QR perpendiculars upon OX, MP respectively.

Then LQPR = 90° – + RQP = 4 RQY = . XOQ= A.

[Or : RP, QP being perpendiculars upon 0X, OY contain the same angle as OX, OY.]

MP MR-PR NQ PR (1) sin (A - B) = sinx0Z =

OP OP OP OP = [introducing required hyps. under the numerators]

NQ OQ PR

PR PQ
OQ OP

OP PQ OP
= sin NOQ.cos POQ-cos QPR.sin POQ
=sin A.cos B - cos A. sin B.

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OM ON + NM ON QR (2) cos (A B) = cos XOZ =

OP OP

OP

OP = [introducing required hyps. under the numerators]

ON OQ

OQ QR PQ

OQ OP' PQ' OP
=cos NOQ.cos POQ + sin QPR. sin POQ

= cos A. cos B+ sin A. sin B.
Since the triangles NOQ, RPQ are similar,

PR PQ ...

tan B, ON OQ

MP NQ - PR (3) tan (A - B).= tan XOZ

OMON + QR
NQ PR

ON ON tan A tan B
ON QR

PR 1 + tan A tan B ON PR ON 331. We may combine the above two figures into one, if we dash the letters Z, P, M, R in the second figure. Thus

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OP

In OY take any point Q; through Q draw PQP perpen

' dicular to OY, cutting OZ, OZ' in P, P.

Draw QN, PM, PM' perpendiculars on OX ; and RQR' perpendicular to QN, cutting PM, P'M' in R, R'. Then

OP = OP', RQ=QR, RP= P'R'.
Also LQP'R' = 4QPR = 90° – LRQP= 2 OQR=A,

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MP MP sin (A + B) + sin (A - B) = sinx0Z+ sin XoZ

OP ОР" (NQ + RP)+(NQ-P'R') 2.NQ_ 2. NQ OQ

2 sin A cos B. OP OP OQ

MP MP sin (A + B) – sin (A - B) = sin XoZ– sin XoZ.

OP OP' (NQ+RP)-(NQ-P'R') 2.RP 2. RPQP

2 cos A sin B. ОР OP QP OP

ОМ" OM cos (A B) + cos (A + B) = cos XOZ' + cos XOZ =

OP

OP (ON+QR') + (ON - RQ) 2.ON 2.ON OQ

= 2 cos A cos B. OP OP OQ OP

OM OM cos (A – B) – cos (A + B) = cos XOZ'

cos XOZ=

ОР

OP (ON + QR')-(ON - RO) 2.RQ 2. RQ QP

= 2 sin A sin B. OP

ОР QP OP

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332. Taking the angle XoZ to be P, and XOZ to be Q in the above figure, from Art. 116, we see that

XOY=} (XOZ + XOZ") = } (P+Q) and

YOZ=} (X0Z-XOZ') = 3 (P-Q).
Hence the above formulæ take the form

sin P+ sin Q = 2 sin } (P+Q) cos } (P-Q),
sin P - sin Q = 2 cos } (P+Q) sin } (P-Q),
cos Q + cos P= 2 cos } (P+Q) cos 1 (P -Q),
cos Q- cos P = 2 sin 1 (P+Q) sin 1 (P - Q).

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