Mathematics
Senior High
θ-π/6=π/3,5/3
になる計算法を解説してほしいです！
(2)2005 (8
6
π
)=1より。
cos (0 - 7/7 ) = 1/1/1
COS
・①
0≦02のとき,
π
* ≤0-1 <117
T
y
I
2
→x
0
6
であるから,この範囲
で①を満たすのは, 0
-
よって8=
π
π
6
1130
π +
536
π
11
π
66
π
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