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is the half of the chord of twice that arc. For if we prolong BD to F, BD will be the half of BF, the chord of the double arc BA.

Since the sine of an arc is half the chord subtended by twice that arc, it evidently follows that the sine of an arc of 300 is half the radius.

5. The definition of the sine also shews that the sine of any acute angle BCA is the same as that of the obtuse angle a CB=180° - angle BCA, which is called its supplement.

6. The cosine of an obtuse angle is always negative. The same is also the case with its tangent cotangent, secant, &c.

7. We easily perceive that the sines increase from 0°, where the cosine is equal to radius till 90°. The sine then equals radius, and the cosine vanishes. The sines then decrease from 90° to 180°, where they become nothing, while at that point the cosine equals the radius taken negatively.

8. We may also observe that the tangents and secants increase from 09, where the tangent is zero, and the secant equal to radius, till 90°, at which point the tangent and secant become equal, parallel, and consequently infinite. The cotangent is then zero, and the consecant equal to the radius.

From 90° to 180° they continually decrease; but are negative. When they arrive at 180°, the tangent vanishes, and the secant is equal to the radius taken negatively. At this same point the cotangent and cosecant become equal, and negatively infinite.

9. The tangent of 45° is equal to radius, as also the colangent. For then the right-angled triangles CEM, CTA became equal and isosceles.

10. These things premised, let the arc BA=A; (preceding figure) the radius CB=CA=1 (which supposition we shall make in future, in order to render the calculations more simple), because of the right angled and similar triangles CBD, CGB, CTA, CEM, we shall have the following proportions and equations.

I. CD? +BD’=CB? ; or sin A+cosA=I=sin? B+cos? B (B being any other arc). Therefore

Sin A+sin B: cos B+cos A:: cos B-COS A : sin A-sin B.
II. CT?ATTAC?.

......or sec? A-tang? A=1. III. CE-CM2_EM?....

......or 1 = cosec? A-cot? A=sec A

tang ?A:

IV. CD: BD::CA: AT....or cos A : sin A:: 1 : tang A.

sin A Therefore sin A=cos A x tang A....or cos A

and tang A= sin A

tang A

COS A

tang A

V. CB: CD :: CT: AT....or 1: sin A :: sec A: tang A.

1 Therefore sec A

sin A VI: CG: GB:: CE: EM....sin A: cos :: 1: cot A.

1 Therefore cot A=

and cot A x tang sin Atang A

cos A

cos A

11. Let it now be required to determine the sine and cosine of the sum of two given arcs AB and BE.

Call s the sine BD of the arc AB; c its cosine; $ the sine EG of the arc EB; c its cosine; and lastly, the cosine required=CF

B cos (AB+BE), and y the sine EF=sine (AB+BE).

H This being premised, because of the similar

DA triangles CBD, CFH, EGH, we shall have

CD (c): CF (1) :: CB (1): CH=:: BD ($): FH=**
Again,

'C 2y-Therefore s's-cc-x and s'=cy-9x. The first equation gives x=cc-ss', and substituting this value in the second, we have (observing that 1-5=c%) y=c+s'c. In general therefore, A and B being any two arcs, we have

sin (A+B)=sin A cos B+sin B cos A

cos (A+B)=cos A cos B-sin A sin B 12. Let (A+B=C, and by substitution we shall have

sin C= sin A cos (C-A)+cos A sin (C-A)

cos C=cos A cos (C-A)-sin A sin (C/A) And treating sin (C-A) and cos (CRA) as unknown quantities, we shall find

Sin (C-A)=sin C cos A-sin A cos C, and cos (C-A)=cos C cos A +sin A sin B. Therefore in general,

Sin (A--B)=sin A cos B-sin B cos A

Cos (A-B)=cos A cos B+sin A sin B. Making A=B, we have sin 2 A=2 sin A cos A ; and cos 2 A =cos? A-sin? A=2 cos? A-1 (by writing for sin? A, its value 1-cos? A). It is therefore easy to obtain the sine and cosine of twice any arc of which we already know the sine and cosine.

With the same facility we may find the sine and cosine of the half of any arc.

For if we make 2 A=C, we shall have sin C= 2 sin 1C cos C, and cos C+1=2 cos? {C. Therefore cos C= 1+cos C, Sin C

sin C and

or sin įC= 2 cos 0

7 2 (1+cos C)

2

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cos C

2

2 (1+cos C) But as these formulas suppose that the gines and cosines are already known, before we proceed farther we must learn to calculate them.

13. And first, it is clear that if we calculate the sines of all the arcs comprized in the quadrant or quarter of a circle, from 1" : to the arc of 90°, we shall be in possession of all the sines from 1" up to the arc of 180°. (See 5).

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Now from 180° to 360° the since are the same, as from 0e to 180, excepting the sign which is negative. Therefore the computation of the sines is reduced to those sines contained within the quadrant.

Again it is clear that the cosines may easily be determined from the formula,

Cos A=v(i-sin A). We shall therefore apply ourselves only to the calculation of the sines.

It will hereafter be shewn that if the radius be 1, the arc of 90°. will be represented by 1.570796326794896, &c.; therefore the arc of 1" is 0.000004848, &c. parts of the radius; and as so small an arc does not sensibly differ from its sine, we may assume 0.000004848, &c. for the sine of the arc of l". Double, triple, &c. this decimal fraetion, and we shall have the sine of 2", 3", &c.

We might have calculated the sines - 2", and then that of 3", &c. by the formulas sin 2 A=2 sin A and sin (A+B)=sin A cos B+sin B cos A ; but it was fou . che difference between such small arcs and their sines was so inconsiderable, that any of these arcs might be taken instead of their respective sines.

Ascending in this manner from seconds to minutes, and continuing the calculation from minutes to degrees, by means of the two preceding formula, we arrive at the sine of 30%. As this sine is known to be equal to half the radius, it serves to verify all the anterior calculations, and we shall then have all the sines from one second up to that of 30°. But not to swell the size of the Tables beyond all bounds, it is usual to insert in them only the sines of minutes and degrees. 14. Suppose now that A-30°, we shall have

Sin (30- +B)=sin 30° cos B+cos 30o sin B. But
Sin 30o=1, and cos 30o=V1-I=I3. Therefore
Sin (30° +B)=sin B V 3+} cos B. And
Sin (30°–B)= cos B-1 sin B v 3. Therefore

Sin (30° +B)=sin (30°-B)+sin B. v 3. Hence it follows that if we know all the sines from (o to 30°, we easily obtain all those between 30, and 600.

15. This being premised, let A-60°, and we shall have sin (60° +B)==cos B V 3+ sin B, and sin (60%-B)= cos B V3 - sin B. Therefore

Sin (60° +B)=sin (60°-B)+sin B. For example, sin 66" --sin 540 + sin 6o-If therefore we know the sines of all the arcs between 300 and 60°, we shall immediately obtain all those between 60° and 90°, which will complete this sort of calculation.

16. Resume the two formulas sin (A+B)=sin A cos B+-sin B cos A, and sin (A-B)=sin A cos B-sin B cos A, and add them together, and we shall have

Sin A cos B-, sin (A+B) + sin(A - B). Subtract the second from the first, and we shall find

Sin B cos A= sin (A+B)- sin (A-B).

COS

COS

COS

Perform the same operations on the other two formulas cos A+B)=cos A cos B-sin A sin B, and cos (A-B)=cos A cos B +sin A sin B, and we shall deduce

Cos A cos B= cos (A+B) + cos (A-B)

Sin A sin B=cos (A+B) - cos (A+B) These four last formulas are useful when we wish to transform products of sines into simple sines. The four following forms enable us to substitute for the sums or differences of sines, the product of other sines, in order that the logarithmic calculus may apply to them. 17. Let A+B=P, A–B=Q, we shall have A=P+Q, and

2 B=P-Q. Therefore from the preceding article

P+Q

P-Q
Sin P+sin sin

2

2

P-Q P+Q
Sin P-sin O=2 sin

2

2
P+Q

P-Q
Cos P+cos Q=2 cos

2

2
P+Q

P-Q
Cos Q-cos P=2 sin sin

2 18. Suppose that in the two first of the foregoing formula P= 90°, and in the two last that Q=0; and we shall have 1 +sin Q=2 sin (45° +Q) cos (45° —Q)=2 sin? (45° + FQ) 1-sin Q=2 sin (45°

-Q) cos (45° +10=2 sin (459-10) — 2 cos? (45° +}Q)= coversed sine of Q

1+cos P=2 cos? } P

1-cos P=2 sin? | P-versed sine of P. 19. Divide the fornulas of No. 17, one by another, and we shall have

P+Q

sin
Sin P+sin Q

2
2

P+Q

tang Sin P-sin Q

2 P+Q P+Q

sin 2.

2 P+Q

tang P-Q

2 cot 2

P-Q
tang.

2
And in like manner,
Sin P+sin Q
P

P-Q
Cos P+cos Q.

2

cos P+cos Q Sin P+sin Q

sin P-sin Q

P+Q

; Cos Q-cos P

2 cos P +cos Q Cos P+cos Q P+Q P-Q

cot Cos Q-cos P

2

P-Q

COS

х

COS

= tang P+Q; sin P-sin Q

= tang P_Q = co bude

= cot

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xcot

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20. In like manner dividing one by another, some of the formula of No. 18, wo find 1 +sin Q_sin(459 +1 Q)_sin (45°+; Q)

45( =tang' (459+Q) 1-sin Q sin? (45°

Q cos" (45° +*Q) 1 +cos P cog* P

=cot } P 1—cos P sin? } P 1+sin P

sino (450+} Q) 1+cos P cos' } P 1-sin Q coversin Q_sin(45°—Q) 1-cos Q versin Qsin} Q

21. Resuming the values of sin (A+B), sin (A-B, cos (A+B), cos (A—B), we deduce from them

sin A cos B sin B cos A Sin '(A+B)_sin A cos B+sin B cos A

sin A sin B Sin (A-B) sin A cos B-sin B cos A

sin A cos B sin B cos A sin A sin B

sin A sin B cos A

1

1

+ sin B

sin A cot B+cot A_tang B'tang A
cos A cot B-cot

A1 1
sin B
sin A

tang B tang A tang A+tang B tang A-tang B

sin A sin B+

cos B

+

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cos B

+

cos A
sin A

ein (A+B)_ sin A cos B+sin B cos A _sin B
cos (A-B) cos A cos B+sin A sin B cos A cos B

+.2

sin A sin B cot B+cot A tang A +tang B

1+cot B cot A 1+tang A tang B sin (A—B) sin A cos B—sin B cos A cot B-cot A cos (A + B) cos A cos B-sin A sin B cot B cot A-1 tang A-tang B 1-tang A tang B cos (A+B)_cos A cos B-sin A sin B cot B-tang A cos (A-B) cos A cos B+sin A sin B

cot B+tang A 1-tang A tang B__cot A-tang B

1+tang A tang Bocot A +tang B sin (A+B)

tang A +tang B cot A +cot B = tang (A+B)= cos (A+B)

1-tang A tang B cot A cot B-1

1 Therefore, cot (A+B=

1-tang A tang B

tang(A + B)tang A +tang B cot A cot B-1 cot A +cot B

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