93 
TRIGONOMETRY. 
10. The tangent of half a right angle is equal to the 
radius. 
11. The sine is equal to the cosine of half a right angle, 
and the square of each is equal to half the square of the 
radius. 
12. The cosine of an angle equal to one-third of two right 
angles is equal to half the radius. 
13. The square of the sine of an angle equal to one-third 
of two right angles is equal to three-fourths of the square of 
the radius. 
14. The sine of an angle equal to one-third of a right 
angle is equal to half the radius. 
15. The square of the cosine of an angle equal to one- 
third of a right angle is equal to three-fourths of the square of 
the radius. 
16. In a triangle the sides are to each other as the sines 
of their opposite angles. 
17. In a triangle the tangent of half the sum of any two 
angles is to the tangent of half their difference, as the sum 
of the opposite sides is to their difference. 
18. In a triangle the cosine of any angle multiplied by 
twice the product of the sides which contain it, is equal to 
the sum of the squares of those sides diminished by the 
square of the third side. 
19. In a right-angled triangle given the two sides contain¬ 
ing the right angle, the triangle itself may be determined. 
20. In a right-angled triangle given the side opposite the 
right angle and one of the remaining sides, the triangle itself 
may he determined. 
21. In a right-angled triangle given the side opposite 
the right angle, and one acute angle, the triangle itself may 
be determined. 
22. In a right-angled triangle given either side about the 
right angle, and either acute angle, the triangle itself may 
be determined.. 
23. In any triangle given two sides and the included angle, 
the triangle itself may be determined. 
24. Given two angles of a triangle, and the side opposite 
either of them, the triangle itself may be determined. 
25. Given two angles of a triangle, and the side between 
them, the triangle itself may be determined. 
26. Given the three sides of a triangle, the triangle itself 
may be determined. 
27. Given two sides of a triangle, and the angle opposite 
either of them, the triangle itself may be in some cases de¬ 
termined. 
We proceed now to the proofs, which will be numbered 
in the same way. 
1. Let a c d (fig. 1.) beany angle represented by a. Then 
if we use r to represent a c the sec. 3 a — lan 3 a + r 3 , 
the proof being exactly the same as in 111, in the article 
Geometry, where it was shewn that in every right-angled 
triangle the square of the side subtending the right angle— 
the sum of the squares of the two sides containing that 
angle. 
2. Let acd (fig. 1.) be any angle represented by a. Then 
tan a X cotan a — r 3 . For the angles hac, acb, cbk, 
being all right angles, ah is parallel to cb, and ac to bk. 
Now (by 95 Geom.) the angle ach nr the angle ckb, and 
the angle ahc — the angle hcb. Therefore, (by 99 Geom.) 
ah : bc :: ac : bk, . •. (by 162 Geom.) ah x bk — bc x ac 
— ac 3 . Q.E.D. 
3. Let acd (fig. 1.) be the angle represented by a. Then 
R 3 — sin 3 a -f cos 3 a. For by 111, Geometry, CD 2 — dr* 
+ GC 2 . Q.E.D. 
4. Let acd (fig. 1.) be the angle, represented by a. 
Then, - A - — which is plainly deducible from Ge- 
cos a r r J 
ometry, 98. 
Ohs. Hence, if the complementary angle bcd be called 
b, it follows by this Art., that 
sin b tan b 
cos b r ’ 
i. e. As the sine, cosine, and tangent of b are respectively 
the cosine, sine, and cotangent of a, 
Voe, XXIV. No. 1629. 
cos a cotan A 
sin A R 
5. Let acd (fig. 1.) be the angle, represented by a. 
Then sec a X cos a r= r 3 . For by 99 Geom., ch: cd :: 
ca : cg ; . • . by 162 Geom.,* ch X cg — cd x ca — ca 3 
Q. E. D. 
Hence it follows that, representing the angle bcd by b, 
sec B X cos B — R 2 ; 
i . e . as the secant and cosine of b are respectively the co¬ 
secant and sine of a, 
cosec a X sin A — R 3 . 
It will easily be perceived that all the above definitions of 
goniometrical lines are applicable, whatever be the magni¬ 
tude of the given angle, although we appear to have con¬ 
sidered those only, such as acd, which are less than a right 
angle. 
Thus in fig. 2, if acd' be the given angle, ah' will be its 
tangent, according to Def. I.; because ah' is “ that portion 
of the geometrical tangent intercepted between the two sides” 
ca, cd', (the latter being produced backwards through c, 
in order to meet the geometrical tangent at h'). Also bcd' 
being the difference between the angle acd' and a right 
angle, is the complement of the former, by Def. III., and 
therefore bk' is the cotangent of acd'. 
Again: ch' is the secant of acd', according to Def. II.; 
and ck', its cosecant. 
Finally: d'g' is the sine of acd' cg' its cosine, ag' 
its versed sine, according to Def. IV. and Obs., and 
Def. V. 
And not only may the definitions given above be extended 
to the caess of such angles as acd', but likewise the asser¬ 
tions made hitherto. 
Thus: 
CH 3 — AH 3 -F AC 2 , 
( 1 ). 
ah' : BC :: ac : bk', 
( 2 ). 
d'g' 3 -J- cg’ 3 — R 2 , 
(3). 
that is, sec 3 acd' — tan 3 acd' + r*. 
tan acd' x cotan acd' — r 3 . 
d'g' : ah' : : CG': AC, 
DI : bk : : ci: cb 
sin 3 acd' + cos 3 acd' — r* 
sin acd' tan acd 1 
cos ACD’ 
cos acd' 
R 
cotan acd' 
.(Art. 4.) 
sec acd' x cos acd' — r 3 . 
cosec acd' x sin acd' — R 2 . 
sin ACD' R 
(Obs. 4.) 
ch' : cd : : ca cg', 
(5). 
ck' : cd' : : cb : ci 
(5. Obs.) 
6. This is evident; for if acd be anya ngle, then a'cd is its 
supplement, and these angles have both the same sine dr. 
7. Let acd (fig. 3.) be any angle, and let the arch dad' be 
twice the arch da, then the sine of acd is equal to half the 
chord dd'. 
Draw cd'. Inasmuch as the arch ad — the arch ad', the 
angle acd — the angle acd', by 99 and 100 Geom. Like¬ 
wise in the triangles acd, acd', the side cd — cd', and the 
side ce is common. Therefore, the side de — ed'. But as 
the angle ced = ced', de is perpendicular to ca, and . •. 
is the sine of acd (Def. IV.): hence sin acd — q.e.d. 
8. Let Acp (fig. 4.) be any angle represented by a ; and 
let dcd' —2a. Then r x sin 2a —2 sin a x cos a. 
Join d'd, and draw dg perpendicular to cd'. As in the 
preceding ced' is a right angle, and . •. — to dgd'. Like¬ 
wise the angle dd'c is common to the two triangles dd'g, 
ed'c ; and . •. these triangles are equiangular to each other. 
Consequently, by Prop. 4, Euclid, cd' : dd' :: ce : cg ; 
but “ if 4 right lines be proportionate, the rectangle under 
the extremes is equal to the rectangle under the means.” . •. 
cd' x dg — dd' x ce — 2de x ce. But cd' is the radius, 
dg is the sine of dcd' or twice acd, de is the sine of acd, 
and ce is the cosine of acd ; hence R X sin 2a — 2 sin A 
X cos a. q.e.d. 
2 B 9. Let 
