T E I
300
TRI
Sec. 45
1
cos. 45
=]> = ^
n/2
60°
Sin. 60 — cos. 90 — 60 — cos. 30 =
V3
2
Cos. 60 =: sin. 90 — 60 = sin. 30 = —.
Tan. 60
Sec. 60 =
\/3
= jl = ^ 3 .
cos. 60 1
1
cos. 60
= r = 2.
120 °
Since the sine of an arc = sine of its supplement, and thecosine of an arc = — cosine of its supplement;
—
2
1
Sin. 120 = sin. 60 r=
Cos. 120 = cos. 60 = —
n/3
2
2
Tan. 120 =
Sec. 120 =
sin. 120cos. 120
= — n/3;
cos. 120
= +
The above, as we observed, are calculated for a circlewhose radius = 1, but if we require their value in a circleof any other radius, it can be readily obtained, for we havethis rule to transform formulas, computed to a radius unity,into others computed to a radius (r) ; write, for sin. a,
cos. a., &c.
sin. a cos. A.
&c., and the values of the sin. a,
and cos. a, &c. so found, will be in terms of the radius (r).
Also if we know the numerical values of sin. A, and cos. a,to radius unity, and we wish to have them to a radius (y),we must multiply the numerical value by (r) ;
^ p n
For -— = sin. A ;.*.»» = r sin. Ar
or sin A to radius r = r x sin. a to rad. = 1.
Ex. Find the numerical value of sin. 30 to rad. 10.000.
sin. 30 to rad. unity = -i- = .5 ;
Ji
.% sin. 30 to rad. 10.000 = 10.000 X -5 = 5000.
We may here observe, that in any right-angled triangle,
altitude
the sine of one of the acute anglesbase
cosine —
hypothenuse
and the tangent =
hypothenuse*altitude
the
base
r ifc s<in ,e
The sine and cosine of any^ two unequal arcs jv Q f the
J. /tC OLft O tl/OC./0O l JJ i/ivy v.. - v
circle being given; to find the sine of any multiplelesser arc. . ^
Let e k be any arc of a circle, which being dividedtwo arcs, e f the less and F k the greater, draw the raCl11 . gf c, and k c ; make f d equal to f e, and join D E cU c jc,c f at l : draw E m, t o, f s, and d h, perpendicular t0 ^cutting c K in M, o, G, and H : parallel to C K drafl^ Ecutting l o at s, and l n cutting e m at n : then let r —• ^= C f = C k, let the arc f k be called b, and the arc $ cQcalled a ; then f g is the sine of b, ct the cosine of A ’^ eS)cosine of b, and e l the sine of a : then, by similar tnan 0C f : f g ; : C 1 : l o, that is,
. . . sin’ b x cos’ A
r : sm b ; : cos A:io= -•
r
Again, by similar triangles,
C f : c G :: el:en, that is,
' , . , cos’ B x sin’ A
r : cos B : : sin’ a : E n = --•
r
, jg f
Now E M = sin’ (A + B) = NM + EN = I'°-rtherefore, by addition,
. , . . sin’ b X cos’ a 4- cos’ b x sin’ a
sin (b — A) = ---
v r
Again, p h = sin’ (b — a) = lo — en;therefore, by subtraction,
... . sin’ b X cos’ A — cos’ b x sin’ A
sm (b — a) = —--
I th 11 ® *
These formulas may be simplified by putting r =
Form. 1. sin’ (b + a) = sin’ b X cos’ A + cos^’ B ^ . n > a;Form. 2. sin’ (b — A) = sin’ b x cos’ A — cos’ B Xadd these formulae together, and we obtain, » \ ;
Form. 3. sin’ (b a) + sin’ (b — a) = 2 x sin’ b X ®_ ^
therefore, sin’ (b + a) = 2 X cos’A + sin’ B — sin f
Let B = (w — 1) a ; then will _g\ A,
sin’ n A = 2 x cos’ a X sin’ (n — 1) a — sin’ (»which is a general theorem. asivelf
Now in this general theorem, let n be put succequal to 1, 2, 3, 4, &c., and we obtain,
s’ 1 A = s’ A.
s’ 2 A = 2 X c’ A X s’ A.
s’ 3 A = 2 X c’ A X s’ 2 A — s’ A.
s’ 4 A = 2 X c’ A x s’ 3 A —T-. s’ 2 A.
s’ 5 a = 2 x c’a X s’4 a — s’3 A, &<=• an do’
where s’ signifies the sine of the quantity that fob 0 ’the cosine of the quantity that follows it.
a
In the last article, put 2 a = a; .-.a = •
, sin. a
. U U
2 sin. — . cos. —;
4 “
d 2 &
and cos. a = 1 — 2 sin. 2 — = 2 cos. '
1,
. of 3°
ar c
in
by which formulas we find the sine and cosineterms of the sine and cosine of half the arc. ^ui t*’ e
A great variety of formulas may be deduced t .general formulas ; the following are the most imp 0Adding 1 and 2 together—
sin. (a -j- b) + sin. (a — b) = 2 sin. A c °
And similarly, also, we have—
cos. (a + b) cos. (a — b) = 2 cos.
sin. (a + b) — sin. (a — b) = 2 sin.
cos. (a — b) — cos. (a + b) = 2 sin.
A cosB cosA sin