TEI
501
TEI
Let A + b = m, and (a — b) = n ;
m + n, , m — n ,a = —--and b = —-—, and
. „ . m + n m — n
sin. m + sin. n = 2 sm. —-—. cos. —-—
_ m + n m — ncos. m + cos. n = 2 cos. —^—. cos. —^—
. m — n m + nsin. m — sin. n — 2 sm. —-—. cos. •—-—
. m + n . m — ncos. n — cos. m = 2 sm. —^— sin. —%—•
Again—
since sin. (a + b) = sin. a cos. b + sin. b cos. aand sin. (a — b) = sin. a co s, b — sin. B cos. Atherefore,
sin. (a + b) sin. (a — b) = sin. 2 a cos. 2 b — sin. 2 b cos. 2 A= sin. 2 a (1 — sin. 2 b) — sin. 2 b (1 — sin. 2 a)
= sin. 2 a — sin. 2 b= (sin. a + sin. b) (sin. A — sin. b)cos. (a + b) = cos. A cos. b — sin. a sin. bcos. (a — b) = cos. A cos. b + sin, a sin, b
cos, (a + b) cos. (a—b)= cos. 2 a cos. 2 b — sin. 2 A sin. 2 b= cos. 2 a (1 — sin. 2 b) — (1 — cos. 2 a) sin. 2 b= cos. 2 a —r sin. 2 B = (cos. A + sin. b) X (cos. A—sin . b).
Again— _
sin. (a + b) = 2 sin. A cos. B — sin. A — bCOS. A + B = 2 COS. A COS. B-COS. A-B.
Hence, if we have three arcs, A + b, a, and a — b, g ivenlr * arithmetic progression, the sine and cosine of A + b may
found from the sines and cosines of A and a — b. Thus,to find sin. a + b, multiply sin. A by 2 cos. b, and sin . A — nV (■— 1,) the sum of the products will be sin. A + b; and t o^ n <l cos. a + B, multiply cos. A by 2 cos. b, and cos. a — bhy (■—. 1 an d add the results together. The ex pression2 cos. b— 1 , is called the scale of relation ; and sin. a + b,8lr >- a, and sin. a — b are called terms of a recurring series.
Since sin. (a + b) + sin. (a — b) = 2 sin. A cos. b,and cos. (a + b) + cos. ( A — b) = 2 cos. a cos b.
For b put a , and for A put na; _
A + B= na + o = n + 1 . aa — b = na — a = n — 1. a;
.% sin. (n + 1 ) a + sin. (n — 1 ) a = 2 sin. n a c^s. a
end cos. n + 1 & + cos. (w — 1 ) a =2 cos. na cos. a.
Fet n= 2; n + 1 = 3, and n — 1 = 1
therefore,
n,3 « + sin. a = 2sin. 2aXcos.a = 2(2 sin.acos. a) cos .a= 4 sin. a cos. 3 a = 4 sin. a (1 — sin. 2 a)
= 4 sin. a — 4 sin. 3 a ;
sin. 3 a = 3 sin. a — 4 sin. 3 aend cos. 3 a + cos. a = 2 cos. 2 a cos. a.
= 2 (2 cos, 2 a — 1 ) cos. a.
= 4 cos. 3 « — 2 cos. a.
cos. 3 a — 4 cos. 3 a — 3 cos. a.
H ^hese formulas for sin. 3 a, and cos. 3 a, give the values ofe _sine and cosine of triple the arc in terms of the sine andof the simple arc.v °i- n. 6 m
To find the sine and cosine of an are in terms of the sine oftwice the arc.
Cos. 2 A + sin. 2 a = 12 sin. A. cos. A = sin. 2 A ;cos. 2 A + 2 sin. A cos. A + sin. sA = 1 + sin. 2 a ;and cos. 2 a — 2 sin. A cos. A + sin. 3 a = 1 — sin. 2 a.
COS. A + sin. A = _+_ 1 sin. 2 a
cos. a — sin. a = J- f 1 — sin. 2 a
cos. A= l j */ 1 + sin. 2 A_+ «/1—sin. 2 a |
sin. A=s i {«✓ 1 + sin. 2 A_+_ n/ 1—sin.2A^
When A is 45 fl , cos. A is 7 sin. a, and we must use
the upper signs; but when A 7 45°, sin. A is 7 cos, a, and
we must use the lower signs.
These values for cos. a and sin. A are termed formulas otverification ; for, by means of them, we are enabled to testor verify the accuracy of results obtained by other processes.Thus, to find sin. 15°,
Sin. 15 = sin. (45 —30) = sin. 45 cos. 30 — sin. 30 cos. 45
1 _ V! — 1
“ 2 ~ 2 n /2 ” 2 V 2
Then, in the formula, let A = 15 ;
/. 2 A = 30 and sin. 2 A = — and l+sin.2A=->l —sin. 2 A=-;
J A
= v/|)
n/3 —12 s/2 ’
the same
result as before.
The utility of the preceding formulas may be shown infinding the numerical values of some arcs.
Sin. (45 + a) = sin. 45 cos. a + sin. a cos. 45
s/2
(cos. a + sin. a).
Cos. (45 + a) = cos. 45 cos. A — sin.45 sin. a
= - 7 = (cos. a — sin. a).s/2
,7 o 1
Let A = 30 ; • cos. A = , and sin A =— ;
2 2
. n /3 + 1 , s/ 3 — 1
sm. 75 = -——and cos. 75 =
2 s/ 2
Next let A = 90 ; cos. A
2 n/T0, and sin. A = 1
\ sin. 135 = —and cos. 135 =-
»/ O k/ O
To find the sine and cosine of 18° and 36°
5 X 18° = 90°; /. 3 X 18° = 90° — 2 x 18°.-.COS. (3 X 18°) = c os. (90° — 2 x 18°)= sin. 2 x' 18°But cos. 3x18 = 4 cos. 3 18 — 3 cos. 18,
Sin. 2 X 18 = 2 sin. 18 cos. 18 ;
4 cos. 3 18 — 3 cos. 18 = 2 sin. 18 cos. 18.
Let 8 = x;
4(1 — * 2 ) — 3 = 2 x, or 4 x 1 + 2x = 1, * 1 1,1 5
+ g + 16 - 4 + 16 16
/, x = —- = sin. 18 = cos. 72,