数学

質問

問７　次の関数のグラフをかけ
(１)y＝２分の一sinθ　(２)y＝２cosθ
問８次の関数のグラフをかけ。また、周期を求めよ
(１)y＝cos2θ　(２)y＝sin2分のθ
問９次の関数のグラフをかけ
(１)y＝sin(θ－６０度)　(２)y＝cos(θ＋３０度)
問１０次の式をrsin(θ＋α)の形に変形せよ
(１)３sinθ－√３cosθ (２)－sinθ＋cosθ
問１１弧度法で表された次の角は、それぞれ何度か
３π

回答No.2

muturajcp

7
(1)
f(θ)=(sinθ)/2
f'(θ)=(cosθ)/2
f((4n-1)π/2)=-1/2
(4n-1)π/2<θ<(4n+1)π/2のとき増加
f((4n+1)π/2)=1/2
(4n+1)π/2<θ<(4n+3)π/2のとき減少
(2)
f(θ)=2cosθ
f'(θ)=-2sinθ
f(2nπ)=2
2nπ<θ<(2n+1)πのとき減少
f((2n+1)π)=-2
(2n+1)π<θ<2(n+1)πのとき増加
8
(1)
周期はπ
f(θ)=cos(2θ)
f'(θ)=-2sin(2θ)
f(nπ)=1
nπ<θ<(2n+1)π/2のとき減少
f((2n+1)π/2)=-1
(2n+1)π/2<θ<(n+1)πのとき増加
(2)
周期は4π
f(θ)=sin(θ/2)
f'(θ)={cos(θ/2)}/2
f((4n-1)π)=-1
(4n-1)π<θ<(4n+1)πのとき増加
f((4n+1)π)=1
(4n+1)π<θ<(4n+3)πのとき減少
9
(1)
f(θ)=sin{θ-(π/3)}
f'(θ)=cos{θ-(π/3)}
f((12n-1)π/6)=-1
(12n-1)π/6<θ<(12n+5)π/6のとき増加
f((12n+5)π/6)=1
(12n+5)π/6<θ<(12n+11)π/6のとき減少
(2)
f(θ)=cos{θ+(π/6)}
f'(θ)=-sin{θ+(π/6)}
f((12n-1)π/6)=1
(12n-1)π/6<θ<(12n+5)π/6のとき減少
f((12n+5)π/6)=-1
(12n+5)π/6<θ<(12n+11)π/6のとき増加
10
(1)
(2√3)sin(θ+(-π/6))
(2)
(√2)sin(θ+(3π/4))
11
540°

回答No.1

noname#157574

数学IIの教科書の三角関数の章を熟読しましょう。