94 
TRIGONOMETR Y. 
9. Let acd be any angle, represented by a; and let 
dcd' — 2a. Then, r x cos 2a zz r 2 —2 sin 2 a. 
By the preceding, 
r X sin 2 a zz 2 sin a x cos a. 
R 2 x sin' 1 2a — 4 sin 2 a X cos 2 a ; 
but, by 3, sin 2 2 a zz r- — cos 2 2a, and cos 2 a zz R- — 
sin 2 a, 
R 2 X(R 2 COS 2 2a) — 4 Si» ! A X (R° S»i s A), 
R* — R 2 X cos 2 2a — 4 sin 2 A X (R 2 — sin 2 a), 
R 2 X cos 2 2a — r 4 — 4 sin 2 a x (R 2 —sin 2 a), 
— R 4 — 4 s/n 2 a X R 2 + 4 sin 4 a, 
zz (R 2 — 2 sin 2 a 2 ), 
rooting both sides, r x cos 2a — R 2 — 2 si?« 2 a. q. e. d. 
If the angle a were divided into two equal parts, we 
should have, by the preceding formula, 
R X cos 2 zz R 2 — 2 sin 2 ; 
i.e. As 2 (A) - A, 
A 
R X cos A — R 2 — 2 sin 2 —; 
in other terms, the rectangle under the radius and cosine of 
an angle is equal to the difference between the square of the 
radius and twice the square of the sine of half the angle ; 
which is, in fact, only the preceding theorem in another 
form of expression. 
By the preceding formula, inasmuch as 
sin 2 ~ — r 2 — cos 2 (by 3), we have 
RXCOSAITR* -2 (r 2 — COS 2 
ZZ 2 COS 2 ^-R 2 , 
•V 
.\ R 2 4- R X cos A 1 = 2 cos 2 — ; 
in other terms, twice the square of the cosine of half an 
angle is equal to the square of the radius, together with the 
rectangle under the radius and the cosine of the angle. 
10. Let acd (fig. 5.) be zz 45°, i. e. half a right angle. 
Then, tan 45° zz R. 
In the triangle hac the angle hac is a right one, ah 
being a tangent of the circle at a ; and hca is given half a 
right angle. Consequently, since by 96 Geom. “ the three 
angles of any triangle are equal to two right angles,” the 
angle ahc must be half a right angle. Hence, by 104 
Geom., the side ah is equal to the side ac; but ah is the 
tan of acd, and ac is the radius, q. e. d. 
11. In the fig. above mentioned, let acd be zz 45°. Then, 
JR 2 
sin 45° zz cos 45°, and sin 2 45° zz cos 2 45° zz —. 
Now draw dg perpendicular to ac; and for the same 
reasons as above, dg zz cg. Likewise, by ] 11 Geom., cd 2 
zz dg 2 4- cg 2 ; that is, cd 2 zz 2 dg 2 . Hence, dg 2 zz cg 2 zz 
CD 2 
But dg is the sine, and cg the cosine of acd, and 
cd is the radius. Q. e. d. 
12. Let acd (fig. 6.) be zz 60°, i. e. one-third of two 
R 
right angles. Then, cos 60° zz -g-. 
Now join ad. The three angles of the triangle acd are 
equal to two right angles, by 96 Geom.; but the angle acd 
is given equal to owe-third of two right angles, therefore the 
remaining angles adc and dac must be together equal to 
/zoo-thirds of two right angles. Wherefore, as cd zz ca, 
the angle adc zz the angle dac, and each of them is equal 
to one-third of two right angles. Consequently, since 
“ every triangle which has its three angles equal, has its 
sides equal," the three sides ac, cd, da, are equal; and 
therefore, by 103 Geom., the side ca is equally divided at 
the point g, by the perpendicular dg, i. e. cg zz C ~. But 
ca is the radius, and cg the cosine of acd, q. e. d. 
13. In the figure above mentioned, let acdzz60°. Then 
Rn! 
sin 2 60°zz — 
4 
Now sin 2 60° 4- cos 2 60zzr 2 by 3; 
R 2 
.•. sin 2 60° 4- — zzR 2 , by the preceding. 
• 9 enn 9 R 2 3r 2 . 
. . sm 2 60° zz R 2 — —■ zz — q. e. d. 
4 4 
14. Let acd, (fig. 7.) be zz30°, i. e. one-third of a right 
angle. Then, sin 30° zz — 
Now since the angle acd is one third of a right angle, the 
angle bcd is two thirds of a right angle; that is, bcd is 
one-third of two right angles. Consequently (by 12 ,) cos 
BCD = 2 ’ k Ut ** ^ ie cosine an a "gl e is sine of the 
complement of the angle,” therefore, cos bcd zz sin acd. 
Hence, sin acd zz^. 
15. In the figure above, let acd be zz 30°. Then 
3n 2 
cos 2 30° ZZ —. 
4 
sin 2 30° 4- cos 2 30° zz R 2 (by 3.) 
.'. — 4- cos 2 30° zz R 2 (by 14.) 
p 2 Qr2 
.'. cos 2 30° ZZ R 2 --7-ZZ — • Q. E. D. 
4 4 
If we consider only such angles as are less than a right 
angle, it follows from the preceeding propositions, that we 
can determine the magnitude of an angle in some cases, by 
knowing certain of its goniometrical lines. Thus, if we 
know that the tangent of an angle is equal to the radius, or that 
its sine or cosine is to the radius as 1: 2, the angle must ne¬ 
cessarily be equal to half a right angle, by 10 and 11, re¬ 
spectively. If we know that the cosine of an angle is equal 
to half the radius, or that its sine is to the radius as 3 : 2, 
the angle must necessarily be equal to one-third of two right 
angles, by 12 and 13 respectively. If we know that the 
sine of an angle is equal to half the radius, or that its cosine 
is to the radius as 3 : 2, the angle must necessarily be 
equal to one-third of a right angle by 14 and 15 respectively. 
But there are certain other angles immediately deter¬ 
minable from their goniometrical lines, which we proceed to 
enumerate. It appears from 6, that the magnitude of an 
angle is not generally determinable from the ratio of its sine 
to the radius, inasmuch as the same sine corresponds to two 
angles, viz., those which are supplementary one to the other. 
But in order to prevent the same ambiguity with regard to 
the other goniometrical lines, mathematicians have recourse 
to the following expedient: All perpendiculars to the di¬ 
ameter aa', fig. 3, such as dg, bc, at the upper side of aa' are 
affected with the sign 4- ; and those, such as ah', cb', at 
the under side of aa', are affected with the sign —. Like¬ 
wise: all perpendiculars to the diameter bb', such as di, 
cg, at the right side of bb', are affected with the sign -f- ; 
and those such as d" T ca', at the left side of bb', are 
affected with the sign —. Thus, ah' is written — ah', 
which distinguishes it from 4- ah, to which it may be equal 
in length; and d" i' is written — d" i', which distinguishes 
it from 4 - di, to which it may be equal in length. It is, 
however, only necessary to write the minus signs to their 
corresponding lines, the others being always understood to 
have the sign 4 -* 
Now, if the radius cd (fig. 8 .) be supposed to revolve 
round the point c, so as to form with its first position ca, 
angles of increasing magnitude, it is plain that the sine dg 
beginning from nothing (when cd coincides with ca), will 
continually augment, as at n, till cd reaches the point r, 
where it will be equal to. cb the radius; thenceforward it 
will continually diminish till it vanishes at the point a' into 
nothing. Hence at the three remarkable points a, b, a' the 
magnitude of the sine will be thus expressed algebraically: 
