288 FORMULZ FOR THE COSINES AND SINES 
4t™cos(m-1)0 
2é 
and ¢? T,,_,=2é" cos (m—2)@. Therefore, by (2), 
a ee 2 cos 9 cos (m—1) @—cos (m—2) gt =2¢" cos m 0. 
Ty, =2t""! cos(m-1)0= =4z™ cos 8 cos (m—1)9, 
§4. To prove (8), we observe, that, by hypothesis, the Law holds 
for the first step, that is, when m=1. Assume that it holds for 
m—1 steps. Then 
4t™ sin (m—1)0 
a OE 
and ké2t,,_,=2d¢" sin (m—2) 6. Therefore, by (4), 
ktg=2tm }2 cos 6 sin (m—1) O—sin (m—2)0} =22” sin m 6: 
which proves the Law universally. 
kt.) —2en sin (Ve — = 42” cos @ sin (m—1)6, 
§5. In equations (7) and (8), m may be negative as well as positive. 
The series (1), starting from the terms T, and a» may be carried not 
only forwards in the direction of the terms T,, T,, &c., but also 
backwards through the terms T_,, T_,, &c.; the relation expressed 
in (2) always subsisting. In fact, by (2), 
Ae bee Mh os Go he eet est No, 
In general, it is easily seen that 
Deg Se Os Ft Dea os0 se ciwes cle saa ee 
Similarl yy: < Gag SS Os 6 ake eeececcshencossethee (10) 
By equating the values of T,, in (9) and (7), we hee | 
T_,, £?2"=2t” cos m 0=2i™ cos (—m 6) 
”. 26°" cos (—m0)=T_,, 
In like manner, by equating the values of 7,, in (10) and (8), we have 
2¢-™ sin (—m 6)=ht_». 
§6. As an instance of the application of the formule which have 
been obtained, we shall now find cos m @ in terms of cos 0, m being a 
positive integer. In (5) substitute for T,, its value in (7), Sachs re- 
‘place ¢ (see §2) by (2 cos 6)"1. Then 
24" cos m 0=1—mt? + CE gy fat —3) — ye (m—4) + &e. 
[2_ = 
1 
2 cos m 0= Ea Foe" + &c. 
= (2 cos 0)"—m (2 cos 0)"~* + ein (5 cos 6)"-+—&e. 
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