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3 tan 2 π：なし sin sin , cos cos , tan tan 6 π = 6π = 2 6 π = 6π

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(1)

次の角のsin , cos , tanθ θ θ の値をそれぞれ求めよ。

７７．三角関数の値

(1) 2 3 2 1 2

sin , cos , tan 3

3π = 2 3 π = − 2 3π = −

(2) 1 3 1

sin , cos , tan

6 2 6 2 6 3

π π π

− = − − = − = −

     

     

(3) sin 0=0, cos 0=1, tan 0=0 (4) sin 1, cos 0

2 2

π = π =

，tan 2

π ：なし

(5) sinπ =0, cosπ = −1, tanπ =0 (6) 3 3

sin 1, cos 0

2π = − 2π = ， 3

tan 2 π：なし

(7) 17 5 1 17 5 3 17 5

sin sin , cos cos , tan tan

6 π = 6π = 2 6 π = 6π = − 2 6 π = 6 π

(8) 3 1 3 1 3

sin , cos , tan 1

4 π 2 4 π 2 4 π

     

− = − − = − =

     

     

 

(2)

(1) 2 θ = 3 π

2 3 2 1 2

sin , cos , tan 3

3π = 2 3π = − 2 3π = −

(2)

6 θ = − π

1 3 1

sin , cos , tan

6 2 6 2 6 3

π π π

− = − − = − = −

     

     

(3) θ =0

sin 0=0, cos 0=1, tan 0=0

(4) 2 θ = π

sin 1, cos 0

2 2

π = π =

，tan 2

π ：なし

(5) θ π=

sinπ =0, cosπ = −1, tanπ =0

(6) 3 θ = 2 π

3 3

sin 1, cos 0

2π = − 2π = ， 3

tan 2 π：なし

(7) 17 θ = 6 π

17 5 1 17 5 3 17 5

sin sin , cos cos , tan tan

6 π = 6 π = 2 6 π = 6 π = − 2 6 π = 6π

(8) 3

θ = − 4 π

3 1 3 1 3

sin , cos , tan 1

4 π 2 4 π 2 4 π

     

− = − − = − =

     

     

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