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Cor. 1. Hence, likewise, the rectangle under the radius and the sine of the difference of two arcs, is equal to the difference of the rectangles under their alternate sines and cosines; or R sinAC'sinAB cosBC-cos AB sin BC.

Cor. 2. If the two arcs A and B be equal, it is obvious that R sin2A=sinA 2cosA.

Cor. 3. Let the arc A contain 45°; then

R sin(459

B)=sin45°(cos B÷sinB) = √ R2 (cosBsinB)

or R sin(45?+B)=R√÷(cosB±sinB).

Cor. 4. Let 2A=C, and, by the second corollary,
R sinC sin C 2cos C.

PROP. II. THEOR.

The rectangle under the radius and the cosine of the sum of two arcs, is equal to the difference of the rectangles under their respective cosines and sines.

Let A and B denote two arcs, of which A is the greater; then R cos(A+B)=cos A cos B-sinA sinB.

For, in the preceding figure, the triangles OBD and OFG being similar, OB: OD :: OF: OG, and OB.OG= OD.OF, and the triangles OBD and CFH being likewise similar, OB : BD :: CF : FH, or GE, and consequently OB.GE BD.CF. Wherefore OB.OG OB.GE= OB.OE=OD.OF-BD.CF; that is,

R cosAC cos AB cos BC-sinAB sin BC.

Cor. 1. Hence, likewise, the rectangle under the radius and the cosine of the difference of two arcs is equal to the sum of the rectangles under their respective cosines and sines; or R.cos AC'= cos AB cos BC+sinAB sinBC. |

Cor. 2. If A and B represent two equal arcs, it will follow, that R.cos2A=cos A-sin A= (cosA+sin A)(cos A-sinA); or, since cosAR2-sinA",

R cos2A=R-2sinA2=2cos A-R2.

Cor. 3. Since, sinA2=‡R(R—cos2A), and

sin BR(R-cos2B); therefore

sin A-sin B = R(cos2B-cos2A),

Cor. 4. Let the arc A be equal to 45o, and
R cos(45° B)=sin45° (cos B—sinB).

Cor. 5. Let 2A=C, and by the second corollary,
R cosC-R-2sin+C2cos C-R'.

PROP. III. THEOR.

Of the equidifferent arcs, the rectangle under the radius and the sum of the sines of the extremes, is equal to twice the rectangle under the cosine of the common difference and the sine of the mean arc.

Let A-B, A, and A+B represent three arcs increasing by the difference B; then

R(sin(A+B)+sin(A-B))=2cosB sinA.

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20F.BD; that is, R(sinAC+sinAC')=2cosBC sinAB.

Cor. 1. Hence, likewise, of three equidifferent arcs, the rectangle under the radius and the difference of the sines of the extremes, is equal to twice the rectangle under the sine of the common difference and the cosine of the mean

arc; or R(sin(A+B)—sin(A—B))=2sinB cosA.

Cor. 2. Hence R(cos(A−B)+cos(A+B))=2cos B cos A, and R(cos(A-B)—cos(A+B))=2sinB sinA.

For OB: OD:: OF: OG:: 20F: 20G or OE+OE, and OB(OE'+OE)=2OF.OD; that is,

R(cos AC'+cosAC)=2cos BC cos AB.

Again, OB BD :: CF : FH :: 2CF: 2FH, or OE'-OE, and OB(OE'—OE)=2CF.BD; that is, R(cos AC-cosAC)=2sinBC sinAB.

Cor. 3. Let the radius be expressed by unit, and the arcs B and A, denoted by a and na; then collectively 2sin a.cos na=sin(n+1)a—sin(n—1)a,

2cos a.sin na=sin(n+1)a+sin(n—1)a,
2sin a.sin na=cos(n-1)a—cos(n+1)a, and
2cos a.cos na= cos(n-1)a+cos(n+1)a.

Cor. 4. Since versB=R-cosB, it follows that
R(sin(A+B)+sin(A—B))=2R sin A—2versB sinA,

and consequently R sin(A+B)=2R șinA—R sin(A—B)– 2versB sinA, or R(sin(A+B)—sinA)=R(sinA-sin(A−B))

-2versB sinA.

In the same way, it may be shown that R(cos(A-B)-cosA) = R(cosA—cos(A+B))—2versB cosA.

Cor. 5. If the mean arc contain 60°, then R(sin(60°+B) -sin(60°-B))=2sinB cos60°, or sin B 2sin30°. But twice the sine of 30° being (cor. 1. def.) equal to the chord of 60° or the radius, it is evident that sin(60° + B)sin(60°-B)=sinB, or

sin(60°+B)=sin(60°—B)+sin B.

Cor. 6. Produce CE to the circumference, join C'I meeting the production of FG in K, and join OK. Since FK is parallel to CI and bisects CC', it likewise bisects IC'; and hence OK is perpendicular to KC', which is, therefore, the sine of half the arc IAC', or of half the sum of the arcs AC and AC', as CF is the sine of half their difference. But (II.21.El.)IC-CC"2=IC.2C'E', or C'K2-CF2 =CE.C'E'; consequently sin2 A B-sin2 BC=sinAC sinAC', or, employing the general notation,

sinA - sinB* sin(A + B) sin(AB) = (2. cor. 3.) R(cos 2B-cos 2A.)

Scholium. By help of this proposition, the sines and cosines of multiple arcs are easily determined; but the expressions for them will become simpler, if, as in cor. 2. the radius be supposed equal to unit. For A, 2A and 3A being three equidifferent arcs,

sinA+sin3A=2cos A sin2A=2cosA 2cosA sin A, or sin3A 4cos A2.sin A-sinA; and

=

CosA+cos3A=2cosA.cos2A=2cosA (2coșA2-1)=

4cos A3-2cos A, or

cos3A4cos A3-3cos A.

Again, since 2A, 3A, and 4A are equidifferent arcs, sin2A+sin4A=2cos A sin3A=8cos A3 sin A-2cosA sinA, or sin4A8cos A3 sinA-4cos A sinA ;

as

cos2A+cos4A=2cos A.cos3A=2cos A(4cos A3—3cosA), or cos4A 8cosA4. 8cos A+1. In like manner, suming the equidifferent arcs 3A, 4A, 5A, the sine and cosine of 5A are found; and this mode of procedure may be continually repeated. To abridge the notation, however, it will be proper to express the sine and the cosine of the arc a, by s and c. The results are thus expressed in a tabular form:

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