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BZ: cz::GF : bg (E) sine of an arc : its cosine::sine double arc : versed

sine supplement of double arc.

BG : GF:: GF : bg (F) versed sine double arc : sine double arc::sine double

arc : versed sine of the supplement of the double arc. (G) By comparing these proportions, with the proportions Prop. XI. and casting out the equal terms, various other proportions may be formed. Thus (W. 103.) it is said

radius : sine::secant : tangent, and (X. 106.) we have

radius: sine::double sine : versed sine double arc, therefore secant : tangent::double the sine : versed sine double the arc, &c. for others.

(H) Corol. I. The rectangle of the radius and the sine of any arc, is equal to double the rectangle of the sine and cosine of half that arc.

For (A. 106) CB : cz::2BZ : GP

Therefore CB X GF=2BZ X cz." (I) COROL. II. The rectangle of radius and half the versed sine of an arc, is equal to the square of the sine of half that arc.

For (X 106.) CB : BZ:: BF : BG
And CB : BZ:: JBF : JBG

That is CB : BZ::BZ : {BG Therefore CB X BG=BZo; now cB is the radius, BG the versed sine of and Bz is the sine of Bi, which is half the arc BF.

(K) COROL. III. The rectangle of radius and half the versed sine of the supplement of an arc, is equal to the square of the cosine of half that arc. For (Z. 106) CB : cz::VF : bg

CB : cz:: ZbF : fbG

CB : cz::cz : LG Therefore CB X Ibg=cz?. Where ce is the radius, by the versed sine of the supplement of the arc BF, and cz the cosine of half the arc BF=cosine of the arc Bi.

(L) COROL. IV. The diameter of the circle is to the versed sine of any arc, as the square of the radius is to the square of the sine of half that arc.

For, (I. 107.) CB X BG=Bz’, multiply by the radius CB,
then CBx BG=CB X BZ
therefore CB : BG:: CB : BZ

that is 2CB : BG:: CB? : BZ? (M) COROL. V. The rectangle of the sine of any arc, and of the cotangent of its half, is equal to double the square of the co

BF,

sine of half that arc. And the same is true, writing tangent and sine, for cotangent and cosine.

COS A , rad
For, if a= any arc, sine of A = (L.104.)and cosa

cot A
rad. sine a

A(M. 104.)
tang A
And rad : cos a:: 2 sine A : sine 2A ;(A. 106.), hence

2 cos A.

rad rad : cos A::

: sine 2A; therefore cot A

2 cos? A sine 2A =

; or sine 2A . cotA=2cos? A, that is

cot A
sine A. cotA=2 cose fa.

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

sine 2A . tang A=2 sine’A, that is sine A. tang la=2 sine, A.

(N) Using the same notation as in K. 104. the following formulæ may be easily obtained, by simple algebraical reductions.* 2 cos A , sine A

2 sine? A 2 cos? A (O) I. Sine 2A=

rad

tang A

cot A 2 rad. sine a 2 rad. Cos A 2 rad? . tang A

2 rad” . tang A sec A

rado +tanga A

sec? A 2 rad?

2 rada , cot A 2 rad, cot A cot A t tang A

rad? + cot? A cosec? A
cos? A
sine? A

rad?

2 sine? A (P) II. Cos 2a =

rad

rad
2 cos? A-rad?,
rada-tang? A

cot A-tang A
rad
rad? + tang? A

cot A + tang A
cot A-rad?
2rad? - sec A

2cos A
rad
rada

rad cot? A+ rad?

sec? A
cosec? A
- 2 rada

cosec A
2 sine A

rad.
cosec? A

2 rad? (Q) III. Tang 2A = rad- tangA

cot A-tang A 2 rad . cos A

sine A
2 rad . COS A .

sine A 2 rad? . cot A rada - 2 sine A

2 cos2 A-rad? cot? A-rad? 2 rad” . tang A

2 rad? . cot A
2 rada - sec? A cosec? A-2 rad2

rad2
tanga A

cot? A - rad (R) IV. Cot 2 A =

2 cot A

cosec A

rado

rad

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sec A

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sec A

..rad

cosec A

2 rad, tang

A

.

2 tang A

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

. rad

sec A

cosec A

cot A-tang A
rad? — 2 sine? A 2 cos? A - rad?

rad2

2 cos A. sine A 2 cos A . sine a
2 rad? — sec? A cosec? A-2 rad
2 tang A

2 cot a
rad . sec? A

sec A . rad (S) V. Sec 2 A=

2 rad?
sec? A

2 cos A
rad3

rad3

rad? + tang?

A

rada 2 cos? A - rad: ~ rad? 2 sine? A rad? tang? A cot A +tang A cot? A+ rad?

rad . cosec? A rad=

rad = cot A — tang A

cot? A-rad?

cosec? A - 2 radz sec A

rad · sec A (T) VI. Cosec 2 A =

2 rad

2 sine a rad . cosec A

rada
rada+tang A

sec? A
2 cos A

2 COS A .
sine A

2 tang A 2 tang A cot A + tang A rad? + cot? A sec? A

cosec? A
2
2 cot A 2 tang A

2 cot A
2 sine? A

2 rad? 2 cos? A (U) VII. Vers 2 A=

rad

rad 2 vérs A. vers sup A 2vers A. (2rad - vers A) 2rad .tang? A rad

rad

rada + tangʻA 2 rad . tanga A 2 rad . tang A

2 rad3 2 rad3 sec? A

cot A+ tang Arad? + cot? A cosec? A 2 rad . sine A sec2 A

rad?

sec A COS A . 2 rad=

rad. cosec A

sec A

2 rad 2 sine? A 2 cos? A (W) VIII. Vers sup 2 A=

rad

rad
2 rad
2 cot? A

2 rad . cot A 2rads

rad = rada+tangA rad? + cot? A

tanga +cot a

sec? A 2cosa.rad 2 cosec? A — 2 rad2 2cosec A-2sinea

rad =

cosec? A 2/verse sup A-rad) 2/rad - vers A rad

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

2 cos A. sine A Covers 2 A = rad sine 2

Arad

rad rad” – (2 cos a sine. A)

and the chord of 2 A = 2 sine A. rad (Y) Also if ba be substituted for a in each of the foregoing

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sec A

rad =

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• rad

cosec A

sec A

2

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2

(B) III. Tang A Fradi — tang ! A

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sec

A

A

2 cos i A. sine A 2 sine į A (Z) I. Sine A=

&c. exactly rad

tang į A as in O. 108.

cos2 Į A-sine L A rad? - 2 sine į A (A) II. Cos A=

&c. rad

rad (see P. 108.) 2 rad?. tang 1 A

2 rad =

&c. rad

2 Ž cott A-tang ta' (see Q. 108.) rada tang? A cota A

rad? (C) IV. Cot a=

&c. 2 tang A

2 cott A (see R. 108.) rad . seco SA

Ž rad =

&c.

? 2 cos A - sec * (see S. 109.)

sec (E) VI. Cosec a= · cosec f Arad . sec 1 A

&c. 2 rad

2 sine } (see T. 109.) 2 sine? } 2 rad?

2 cos? JA (F) VII. Vers as

&c. rad

rad (see U. 109.)

2rada 2 sine? } A 2 cos? I A (G) VIII. Vers sup A=

&e. rad

rad (see W. 109.)

(H) By finding the values of sine ja, cos įa, tang ta, &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.

rad I

=rad A=

sec A --- rad 2 rad

2 sec A rad– (rad . cos A) 2 = + rad + (rad . sine A)

2 rad – verssupa ✓rado-rad. sine A=

=rad 2

2 rad = chord A.

rad + cos a

sec A + rad (K) II. Costa=rad

2 rad

2 sec A rada +(rad. COS A

=ivrad+ (rad. sine A) + į rad2 – (rad . sine A) 2 2rad' - (rad . vers A)

rad . vers sup A 2

2

COS A

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=rad V

rad.vers A

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=radV

V 21

rad

(L) III. Tang ža=rad rad- cos a

=rad V

vers A

V

=rad V

=rad V

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rad

sec A-rad rad + cos A

sec A+ rad rad . sine A rado (rad . cos a)

=rad =cosec A-cot A= rad + cos A

sine A

2 rad - vers sup A 2 rad – vers a

vers sup A rad + COS A

sec A + rad (M) IV. Cot i a=rad

rad-COS A

sec A

rad rad . sine A rad + (rad . cos a) =cot A + cosec A = rad - COS A

sine A 2 rad- (rad vers A)

vers sup A vers A

2 rad - vers sup A 2 rad

2 sec A (N) V. Secia=rad

=rad rad + COSA

rad + sec A 2 tang

2 rad

2 rad Erad

=rad V
sine A + tang A
2 rad

vers supa 2 rad

2 sec A (O) VI. Cosecha=rad

rad - COS A

sec A

rad 2 tang A 2 rad

2 rad =rad

-rad

=rad tanga-sinen

versa

2rad -verssupa And in the same manner the versed sines, coversed sines, chords, &c. of the half arcs may be found.

=rady

=rad

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vers A

=rad

GENERAL PROPERTIES OF $INES, 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 rin parallel to the diameter; join Ao and produce it to n, draw che perpendicular to A0, 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 3 of III.): hence ph is bisected in g; therefore an is the sum of the sines AG and on, and AP their difference. The arcs ao and or are bisected in E and B (Euclid 30 of III.), therefore AF

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