4 arc tan // - arc tan 239
d= arc tan {. ß= arc tan 239 Ło'c,
tan 4α= tan (2α+2α) tan 20 = tan (α+ α)
tan 2d = tan (d+d)
Xand+tand
tand tand
=
144
25
719
=
5
25
tan 4α = tan (2α+2+)
tanza+tanza
(-tanza tanza
(0
1-
(20
119
の値を求めよ.
25
144
>
tan d = 5.
tan (4d-(³)
=
で
(-== CaCE, -I - p² = ²²
fet
考える
=
dan 40-tan
1+tan 40- tane
(20
119
(+
(20
L
119 239
28680-119
28441
(+
28561
28441
1
239
120
28441
28561
28441
tan (4α-B) = 1 + 4/
:. 4 arc tan = = - arctan 525/2 = 7
てC
239
4
tan ß = 23/9 [188.
となる
4α-ß.
(答)
Gge