Mathematics
Senior High
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A
256 次の値を求めよ。
(1) sin 17/0
(4) COS
元
5
(2) sin 127
7
(5) tan
12
(3) cos
3/8
T
TC
(6) tan
8
(4) cos-
8
13
12
=
4
√√√2-√2
T = COS
2
5
4+6
√4
問題 247
ゆえにπ
5
=COS
COS-
-
4
6
すなわち
√3
√2
=
2
2
2
[TAS] 6-√2
=
4
7
(5) tan_
=tan/.
π
π
12.
田
5
sin 4 sing
.
1
2
√6+√2
4
T
S
-61
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