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that are.

And the same is true, writing tangent and sine, for cotan

gent and cosine.

For, if A any arc, sine of A=

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tang A

And rad: cos A: 2 sine A: sine 2A (A. 100), hence

rad: cos A::

sine 2A

2 cos A. rad

cot A

2 cos2 A

cot A

:sine 2A; therefore

; or sine 2A.cot A=2 cos2 A, that is

sine A.cot A2 cos? A.

Again, by substituting the value of the cosine of A in the second term of the first proportion, we shall obtain

I

sine 2 A.tang A=2 sine2 A, that is sine A.tang A2 sine2 A (N) Using the same notation as in K. 98, the following formulæ may be easily obtained, by simple algebraical reductions*. 2 cos A sine A 2 sine2 A 2 cos2 A

(0) I. Sine 2A=

2 rad. sine A

sec A
2 rad2

cot Atang A

2 rad. cos A

cosec A

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2 rad2. cot A
rad2+cot2 A

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2 rad2. tang A 2 rad2. tang A

rad2+tang2 A

2 rad2.cot A

sec2 A

cos2 A

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rad

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rad tang A

cot Atang A

rad =

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* Emerson's Trigonometry, 2d edit. Prop II. Scholium.

cot A

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rad

(X) The coversed sine may be expressed in terms of the rest by substituting its value in any of the above forms.

Covers 2 A = rad sine 2 A rad

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2 COS A sine A

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rad

and the chord of 2 A=2 sine A.

(Y) Also if A be substituted for A in each of the foregoing expressions, we shall have,

(Z) I. Sine

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(see W. 103.)

2 rad2-2 sine2 A2 cos A

(H) By finding the values of sine A, cos A, tang A, &c. from the most convenient of the foregoing equations, the sine, cosine, &c. of the half arc will be obtained in terms of the sine, cosine, &c. of the whole arc, by easy algebraic reductions.

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={√/rad2+(rad.sine^)+1⁄2√rad2—(rad.sine▲)

2 rad2-(rad.vers A)

2

rad. vers sup A

(L) III. Tang

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And in the same manner the versed sines, coversed sines, chords, &c. of the half arcs may be found..

GENERAL PROPERTIES OF SINES, TANGENTS, &c. OF THE SUMS, AND OF THE DIFFERENCES OF ARCS.

PROPOSITION XIII. (Plate I. Fig. 2.)

(P) The sum of the sines of two arcs is to their difference, as the tangent of half the sum of those arcs is to the tangent of half their difference.

Let BA and Bo be the two arcs; draw the diameter BX, and OD and AG perpendicular to it. Produce OD to meet the circumference in F, and draw FHN parallel to the diameter; join Ao and produce it to N, draw cne perpendicular to Ao, and at E draw EIK perpendicular to cne, meeting co and CB (produced) in I and K.

Because of is perpendicular to BC, it is bisected in D (Euclid III. and 3.), hence PH is bisected in G; therefore AH is the sum of the sines AG and OD, and AP their difference. The arcs Ao and of are bisected in E and B (Euclid III. and 30), there

P

fore

fore AF is the sum of the arcs BO and BA, and Ao is their difference; BE is the half sum, and OE the half difference; but EK and EI are the tangents of the arcs BE and OE.

PH being bisected in G, ON will be bisected in м, now since Ao is bisected in n, and on in м, nм will be the half of An, therefore

AN AO:: nм: no and

AN: AO::AH: AP per similar triangles.

Hence nм:20:: AH: AP

but nмM: no:: EK: EI per similar triangles; therefore AH: AP:: EK: EI

viz. sine AB+sine OB : sine AB-sine OB ::

Tang (BA+BO): tang (BA-BO). Q.E.D.

(Q) Hence, if the two arcs BA and Bo be represented by a and B, we shall have, sine a+sine B: sine A-sine B: tang (A+B): tang (A-B),

or,

sine A+ sine в tang (A+B)

B

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PROPOSITION XIV. (Plate I. Fig. 2.)

(R) The sum of the cosines of two arcs, is to their difference; as the co-tangent of half the sum of those arcs is to the tangent of half their difference.

Let BA and Bo be the two arcs, as in the preceding proposition, BE their half sum, and OE their half difference.

Draw no parallel to AG, then because AO is bisected in », GD (PO) will be bisected in Q, viz. GQ=QD.

Hence 2cQ=CD+CG the sum of the cosines

and 2GQ CD-CG the difference of the cosines.
By similar triangles

nt (=CQ): ni (=GQ);;nr:na

or 2cQ: 2GQ :: nr: na

But nr: nA:: ES: ET

therefore 2cQ: 2GQ:: ES: ET, where ES is the tangent of the complement of the arc BE, or its cotangent; and ET=EI, is the tangent of the arc OE.

Consequently CD+CG: CD-CG:: cot arc BE: tang arc os.

Q.E.D.

(S) Let the two arcs BA and Bo be represented by A and B, COS A+COS B cot (A+B)

then

=

Σ

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COS B-COS Atang (A — B)

PROPOSITION XV. (Plate I. Fig. 2.)

(T) The sum of the tangents of two arcs, is to their difference

as

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