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By considering the arc AM, and its supplement AM', and recollecting what has been said, we readily see that,

sin (an arc)=sin (its supplement)
cos (an arc)=-cos (its supplement)
tang (an arc)=-tang (its supplement)
cot (an arc)=—cot (its supplement).

S'

D

It is no less evident, that if one or several circumferences were added to any arc AM, it would still terminate exactly at the point M, and the arc thus increased would have the same sine as the arc AM; hence if C represent a whole circumference or 360°, we shall have sin x=sin (C+x)=sin x=sin (2C+x), &c.

The same observation is applicable to the cosine, tangent, &c.

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S

MAT

PA

V

Hence it appears, that whatever be tne magnitude of x the proposed arc, its sine may always be expressed, with a proper sign, by the sine of an arc less than 180°. For, in the first place, we may subtract 360° from the arc as often as they are contained in it; and y being the remainder, we shall have sin x=sin y. Then if y is greater than 180°, make y=180° +z, and we have sin y=-sin z. Thus all the cases are reduced to that in which the proposed arc is less than 180°; and since we farther have sin (90°+x)=sin (90°-x), they are likewise ultimately reducible to the case, in which the proposed arc is between zero and 90°.

XIV. The cosines are always reducible to sines, by means of the formula cos A=sin (90°-A); or if we require it, by means of the formula cos A=sin (90° +A): and thus, if we can find the value of the sines in all possible cases, we can also find that of the cosines. Besides, as has already been shown, that the negative cosines are separated from the positive cosines by the diameter DE; all the arcs whose extremities fall on the right side of DE, having a positive cosine, while those whose extremities fall on the left have a negative cosine.

Thus from 0 to 90° the cosines are positive; from 90° to 270° they are negative: from 270° to 360° they again become positive; and after a whole revolution they assume the same values as in the preceding revolution, for cos (360° +x)=cos x.

From these explanations, it will evidently appear, that the sines and cosines of the various arcs which are multiples of the quadrant have the following values:

sin 0°=0
sin 180° 0
sin 360°=0
sin 540°=0

sin 90°-R
sin 270°-R
sin 450°-R

Cos 90=0
cos 2700
cos 450°-0
cos 630°=0
cos 810°=0
&c.

sin 630°-R

sin 720°=0 sin 810°-R

&c.

&c.

And generally, k designating any whole number we shall have

COS 0°=R
cos 180°-R
cos 360° R
cos 540°-R
cos 720° R
&c.

sin 2k. 90° 0,

sin (4k+1). 90°=R,

sin (4k-1). 90°——R, cos (4k+2). 90°——R. What we have just said concerning the sines and cosines renders it unnecessary for us to enter into any particular detail respecting the tangents, cotangents, &c. of arcs greater than 180°; the value of these quantities are always easily deduced from those of the sines and cosines of the same arcs : as we shall see by the formulas, which we now proceed to explain.

cos (2k+1). 90°=0,
cos 4k. 90° R,

For the radius CA, perpendicular to the chord MN, bisects this chord, and likewise the arc MAN; hence MP, the sine of the arc MA, is half the chord MN which subtends the arc MAN, the double of MA.

THEOREMS AND FORMULAS RELATING TO SINES, COSINES, TANGENTS, &c.

XV. The sine of an arc is half the chord which subtends a double arc.

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N

D

R

E

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The chord which subtends the sixth part of the circumference is equal to the radius hence

;

360° sin or sin 30°R,

12

other words, the sine of a third part of the right angle is equal to the half of the radius.

A

XVI. The square of the sine of an arc, together with the square of the cosine, is equal to the square of the radius; so that in general terms we have

sin 2A+ cos A=R2. This property results immediately from the right-angled triangle CMP, in which MP2+ CP2=CM2.

B

P'

D

R

E

MɅT

It follows that when the sine of an arc is given, its cosine may be found, and reciprocally, by means of the formulas cos A±√(R2-sin2A), and sin A=±√(R2—cos2A). The sign of these formulas is +, or —, because the same sine MP answers to the two arcs AM, AM', whose cosines CP, CP, are equal and have contrary signs; and the same cosine CP answers to the two arcs AM, AN, whose sines MP, PN, are also equal, and have contrary signs.

Thus, for example, having found sin 30°-R, we may deduce from it cos 30°, or sin 60° = √ (R2—¿R2) = √ 3R2={Ř√ 3.

S

PA

XVII. The sine and cosine of an arc A being given, it is required to find the tangent, secant, cotangent, and cosecant of the

same arc.

The triangles CPM, CAT, CDS, being similar, we have the proportions:

CP: PM :: CA: AT; or cos A: sin A:: R: tang A

=

CP: CM:: CA: CT; or cos A: R::R: sec A=

PM: CP:: CD: DS; or sin A cos A:: R: cot A=

R sin A

cos A

R2 cos A

R cos A

sin A R2

PM: CM:: CD: CS; or sin A: R:: R: cosec A

sin A

which are the four formulas required. It may also be observed, that the two last formulas might be deduced from the first two, by simply putting 90°-A instead of A.

From these formulas, may be deduced the values, with their proper signs, of the tangents, secants, &c. belonging to any arc whose sine and cosine are known; and since the progressive law of the sines and cosines, according to the different arcs to which they relate, has been developed already, it is unnecessary to say more of the law which regulates the tangents and secants.

By means of these formulas, several results, which have already been obtained concerning the trigonometrical lines may be confirmed. If, for example, we make A=90°, we shall have sin A=R, cos A=0; and consequently tang 90°=== R2

0'

an expression which designates an infinite quantity; for the quotient of radius divided by a very small quantity, is very great, and increases as the divisor diminishes; hence, the quotient of the radius divided by zero is greater than any finite quantity.

sin

COS

The tangent being equal to R.it follows that tangent and cotangent will both be positive when the sine and cosine have like algebraic signs, and both negative, when the sine and cosine have contrary algebraic signs. Hence, the tangent and cotangent have the same sign in the diagonal quadrants: that is, positive in the 1st and 3d, and negative in the 2d and 4th; results agreeing with those of Art. XII.

The Algebraic signs of the secants and cosecants are readily determined. For, the secant is equal to radius square divided by the cosine, and since radius square is always positive, it follows that the algebraic sign of the secant will depend on that of the cosine: hence, it is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd.

Since the cosecant is equal to radius square divided by the sine, it follows that its sign will depend on the algebraic sign of the sine hence, it will be positive in the 1st and 2nd quadrants and negative in the 3rd and 4th.

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COS

; and cotangent to R. ;

SIL

XVIII. The formulas of the preceding Article, combined with each other and with the equation sin "A+cos 2A=R2. furnish some others worthy of attention.

R2 sin2 A

First we have R2 + tang2 A = R2 + cos2 A R2 (sin2 A+ cos2 A). R4 -; hence R2+tang2 A-sec2 A, a COS 2A cos2 A formula which might be immediately deduced from the rightangled triangle CAT. By these formulas, or by the right-an. gled triangle CDS, we have also R2+cot2 A=cosec2 A.

Lastly, by taking the product of the two formulas tang A= R sin A

R and cot A:

cos A sin A

we have tang A× cot A=-R2, a

"

cos A

formula which gives cot A

R2

We likewise have cot B= tang B'

R tang A

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Hence cot A cot B: tang B: tang A; that is, the cotangents of two arcs are reciprocally proportional to their tangents. The formula cot A × tang A=R2 might be deduced immediately, by comparing the similar triangles CAT, CDS, which CA :: CD: DS, or tang A: R R cot A

give AT

XIX. The sines and cosines of two arcs, a and b, being given, it is required to find the sine and cosine of the sum or difference of these arcs.

Let the radius AC=R, the arc AB-a, the arc BD=b, and consequently ABD=a+b. From the points B and D, let fall the perpendiculars BE, DF upon AC; L from the point D, draw DI perpendicular to BC; lastly, from the point I draw IK perpendicular, and IL parallel to, AC.

F'

C

FK KE P

The similar triangles BCE, ICK, give the proportions, CB: CI :: BE: IK, or R: cos b:: sin a: IK=

sin a cos h

D'

D

L

B

M

A

R
cos a cos b.
R

CB: CI :: CE: CK, or R: cos b:: cos a: CK=·

The triangles DIL, CBE, having their sides perpendicular, each to each, are similar, and give the proportions,

CB: DI :: CE : DL, or R : sin b:: cos a: DL=

cos a sin t.
R

sin a sin l

R

CB: DI:: BE: IL, or R : sin b:: sin a : IL=

But we have

IK+DL=DF=sin (a+b), and CK—IL=CF=cos (a+b).

Hence

sin (a+b)=

sin a cos b+sin b cos a
R
cos a cos b-sin a sin b.
R

cos (a+b)=

C

The values of sin (a-b) and of cos (a-b) might be easily deduced from these two formulas; but they may be found directly by the same figure. For, produce the sine DI till it meets the circumference at M; then we have BM=BD=b, and MI-ID=sin b. Through the point M, draw MP perpendicular, and MN parallel to, AC: since MIDI, we have MN =IL, and IN-DL. But we have IK-IN-MP sin (a-b), and CK+MN=CP=cos (a-b); hence

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