-
268. 次の式を sino, cose, tan 0 のいずれかで表せ。
(1) sin(+47), сos (0-2π), tan (0+67) Bi
(2)* sin(π-0), cos (π-0), tan(-0)
1
268. (1) sin(0+4π)=sin(0+2π×2)=sin0
cos (0-27)=cos {0+2π×(−1)} = cos 0
tan (0+67)=tan(0+2×3) = tan 0
(3) sin
(2) sin(π-0)=sin0
COS...
(π-0) =-cos0
tan (π-0)=-tan 0
COS
5
5
☐(5)
(5) sin(0+2), cos (0-2). tan (0-2)
tan
(3 sin(-0), cos(-0). tan (4-0)
2
2
2
2.4+05003
3
3
(4) sin(0-2x). cos(0+2x). tan (0+2)
(6
T
2
π
2
π
2
(4) sin(0-
-0
0 =sin 0
0
3
= cos 0
cos (0+3
2)=sin(0+
1
tan 0
cos A
= COS
(5) sin(0+x)=
π
・2
= -tan
=COS
π
2
72
= -tan
π
22
tan (0+32 7) = =tan(0-2+2x) =tan(0-7)
2
1
tan 8004
π
2
-2π
-7)=sin(0
cos (0-5) = cos (0-12-27
+2π
π
0 =sin(
=sin(+
0 ==
・+2ヶ
tan (0-2)=tan(0-2-
=cosi -
-2π
0 =sin(
= cos(0-
=sin(0+
9 + 7/7)
70
2
= tan
0 ==
29
1
tan
22
(7)
π
n (0 - 17/7)
12
10
R
cos 6
= cos 0
sin (0+2nn)=sin(
(nは整数)
cos (0+2nn)=cos (
(nは整数)
3tan (0+2nn)=tan 0
(nは整数)
4sin (1-0)=sin
cos (π-0)=-cos
6tan (1-0) = -tan 0
=cos 0
sin
co(-0)-sine
an(-)-tan
-0
0
@sin (0+4)=cos 0
2
cos(-0) = cos 0
12tan (-0)--tan
20
1-0 nia L-0205
(+0niza) arak