基礎数学 No.17 2005. 7. 4
5.3
三角関数の計算公式（解答）
担当：市原問題 41 tanθ= 2, π5θ5 32πのとき, sinθ, cosθを求めなさい. 1 + tan2θ= cos12θ より, cos2θ= 15.
π 5θ5 3π2 より, cosθは負なので, cosθ=− q1
5 =−√1
5 =−√55. tanθ= cossinθθより, sinθ= tanθ×cosθ= 2× −√22 =−2√55.
問題 42 次の値を求めなさい. (1) sin
³
−π 12
´
= sin(−15◦) = sin(30◦−45◦) = sin 30◦×cos 45◦−cos 30◦×sin 45◦
=¡1
2
¢×
³√
22
´
−
³√
23
´
×
³√
22
´
= √42 −√46 = √2−4√6 (2) cos5π
12 = cos 75◦= cos(30◦+ 45◦) = cos 30◦×cos 45◦−sin 30◦×sin 45◦
=
³√
23
´ ³√
22
´
−¡1
2
¢ ³√2
2
´
= √46 −√42 = √6−4√2 (3) tan7π
12 = tan 105◦ = tan(60◦+ 45◦) = 1−tan 60tan 60◦+tan 45◦×tan 45◦◦ = √1−3+1√3
= (
√3+1)×(1+√ 3) (1−√
3)×(1+√
3) = 3+21−3√3+1 = 4+2−2√3 =−2−√ 3 (4) sin3π
8 = sin 67.5◦
0<67.5<90より, sin 67.5◦ =
q1−cos 135◦
2 =
r
1−(−
√2 2 )
2 =
q
2+√ 4 2 =
√
2+√ 2 2
(5) cos9π
8 = cos 202.5◦
180<202.5<270より, cos 202.5◦ =−
q1+cos 405◦
2 =−
r
1+
√2 2
2 =−
q
2+√ 2
4 =−
√
2+√ 2 2
(6) tan7π
8 = tan 157.5◦
90<157.5<180より, tan 157.5◦=−
q1−cos 315◦ 1+cos 315◦ =−
r
1−
√2 2
1+
√2 2
= q2−√
2 2+√
2
=
q(2−√ 2)(2−√
2)
4−2 =p
3−2√
2 =−(√
2−1) = 1−√ 2 (7) sinπ
4sin π
12 = sin 45◦sin 15◦ = −cos(45◦+15◦)+cos(452 ◦−15◦)
= −cos 60◦2+cos 30◦ = −12+
√3
2 2 = −1+4√3 (8) cos5π
8 cos3π
8 cos 112.5◦cos 67.5◦= cos(112.5◦+67.5◦)+cos(112.5◦−67.5◦) 2
= cos 180◦2+cos 45◦ = −1+
√2 2
2 = −2+4√2 (9) cosπ
8 −cos7π
8 = cos 22.5◦−cos 157.5◦ =−2 sin¡22.5◦+157.5◦ 2
¢sin¡22.5◦−157.5◦ 2
¢
=−2 sin(1802◦) sin(−1502 ◦) =−2×1× µ
−
√
2+√ 2 2
¶
=p 2 +√
2
