解
解く y,sin(x+y)+sin(y+z)+sin(x+z)=0
解
y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z,y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
解答ステップ
sin(x+y)+sin(y+z)+sin(x+z)=0
三角関数の公式を使用して書き換える
sin(x+y)+sin(y+z)+sin(x+z)
和・積の公式を使用する: sin(s)+sin(t)=2sin(2s+t)cos(2s−t)=sin(x+z)+2sin(2x+y+y+z)cos(2x+y−(y+z))
2sin(2x+y+y+z)cos(2x+y−(y+z))=2cos(2x−z)sin(2x+2y+z)
2sin(2x+y+y+z)cos(2x+y−(y+z))
類似した元を足す:y+y=2y=2sin(22y+x+z)cos(2y+x−(y+z))
拡張 x+y−(y+z):x−z
x+y−(y+z)
−(y+z):−y−z
−(y+z)
括弧を分配する=−y−z
マイナス・プラスの規則を適用する+(−a)=−a=−y−z
=x+y−y−z
類似した元を足す:y−y=0=x−z
=2cos(2x−z)sin(22y+x+z)
=sin(x+z)+2cos(2x−z)sin(2x+2y+z)
sin(x+z)+2cos(2x−z)sin(2x+2y+z)=0
sin(x+z)を右側に移動します
sin(x+z)+2cos(2x−z)sin(2x+2y+z)=0
両辺からsin(x+z)を引くsin(x+z)+2cos(2x−z)sin(2x+2y+z)−sin(x+z)=0−sin(x+z)
簡素化2cos(2x−z)sin(2x+2y+z)=−sin(x+z)
2cos(2x−z)sin(2x+2y+z)=−sin(x+z)
以下で両辺を割る2cos(2x−z);x=π+4πn+z,x=3π+4πn+z
2cos(2x−z)sin(2x+2y+z)=−sin(x+z)
以下で両辺を割る2cos(2x−z);x=π+4πn+z,x=3π+4πn+z2cos(2x−z)2cos(2x−z)sin(2x+2y+z)=2cos(2x−z)−sin(x+z);x=π+4πn+z,x=3π+4πn+z
簡素化sin(2x+2y+z)=−2cos(2x−z)sin(x+z);x=π+4πn+z,x=3π+4πn+z
sin(2x+2y+z)=−2cos(2x−z)sin(x+z);x=π+4πn+z,x=3π+4πn+z
三角関数の逆数プロパティを適用する
sin(2x+2y+z)=−2cos(2x−z)sin(x+z)
以下の一般解 sin(2x+2y+z)=−2cos(2x−z)sin(x+z)sin(x)=a⇒x=arcsin(a)+2πn,x=π+arcsin(a)+2πn2x+2y+z=arcsin(−2cos(2x−z)sin(x+z))+2πn,2x+2y+z=π+arcsin(2cos(2x−z)sin(x+z))+2πn
2x+2y+z=arcsin(−2cos(2x−z)sin(x+z))+2πn,2x+2y+z=π+arcsin(2cos(2x−z)sin(x+z))+2πn
解く 2x+2y+z=arcsin(−2cos(2x−z)sin(x+z))+2πn:y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
2x+2y+z=arcsin(−2cos(2x−z)sin(x+z))+2πn
以下で両辺を乗じる:2
2x+2y+z=arcsin(−2cos(2x−z)sin(x+z))+2πn
以下で両辺を乗じる:222(x+2y+z)=2arcsin(−2cos(2x−z)sin(x+z))+2⋅2πn
簡素化
22(x+2y+z)=2arcsin(−2cos(2x−z)sin(x+z))+2⋅2πn
簡素化 22(x+2y+z):x+2y+z
22(x+2y+z)
数を割る:22=1=x+2y+z
簡素化 2arcsin(−2cos(2x−z)sin(x+z))+2⋅2πn:2arcsin(−2cos(2x−z)sin(x+z))+4πn
2arcsin(−2cos(2x−z)sin(x+z))+2⋅2πn
数を乗じる:2⋅2=4=2arcsin(−2cos(2x−z)sin(x+z))+4πn
x+2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn
x+2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn
x+2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn
xを右側に移動します
x+2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn
両辺からxを引くx+2y+z−x=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x
簡素化2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x
2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x
zを右側に移動します
2y+z=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x
両辺からzを引く2y+z−z=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x−z
簡素化2y=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x−z
2y=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x−z
以下で両辺を割る2
2y=2arcsin(−2cos(2x−z)sin(x+z))+4πn−x−z
以下で両辺を割る222y=22arcsin(−2cos(2x−z)sin(x+z))+24πn−2x−2z
簡素化
22y=22arcsin(−2cos(2x−z)sin(x+z))+24πn−2x−2z
簡素化 22y:y
22y
数を割る:22=1=y
簡素化 22arcsin(−2cos(2x−z)sin(x+z))+24πn−2x−2z:arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
22arcsin(−2cos(2x−z)sin(x+z))+24πn−2x−2z
数を割る:22=1=arcsin(−2cos(2x−z)sin(x+z))+24πn−2x−2z
数を割る:24=2=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z
解く 2x+2y+z=π+arcsin(2cos(2x−z)sin(x+z))+2πn:y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
2x+2y+z=π+arcsin(2cos(2x−z)sin(x+z))+2πn
以下で両辺を乗じる:2
2x+2y+z=π+arcsin(2cos(2x−z)sin(x+z))+2πn
以下で両辺を乗じる:222(x+2y+z)=2π+2arcsin(2cos(2x−z)sin(x+z))+2⋅2πn
簡素化
22(x+2y+z)=2π+2arcsin(2cos(2x−z)sin(x+z))+2⋅2πn
簡素化 22(x+2y+z):x+2y+z
22(x+2y+z)
数を割る:22=1=x+2y+z
簡素化 2π+2arcsin(2cos(2x−z)sin(x+z))+2⋅2πn:2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
2π+2arcsin(2cos(2x−z)sin(x+z))+2⋅2πn
数を乗じる:2⋅2=4=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
x+2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
x+2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
x+2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
xを右側に移動します
x+2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn
両辺からxを引くx+2y+z−x=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x
簡素化2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x
2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x
zを右側に移動します
2y+z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x
両辺からzを引く2y+z−z=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x−z
簡素化2y=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x−z
2y=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x−z
以下で両辺を割る2
2y=2π+2arcsin(2cos(2x−z)sin(x+z))+4πn−x−z
以下で両辺を割る222y=22π+22arcsin(2cos(2x−z)sin(x+z))+24πn−2x−2z
簡素化
22y=22π+22arcsin(2cos(2x−z)sin(x+z))+24πn−2x−2z
簡素化 22y:y
22y
数を割る:22=1=y
簡素化 22π+22arcsin(2cos(2x−z)sin(x+z))+24πn−2x−2z:π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
22π+22arcsin(2cos(2x−z)sin(x+z))+24πn−2x−2z
数を割る:22=1=π+arcsin(2cos(2x−z)sin(x+z))+24πn−2x−2z
数を割る:24=2=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z
y=arcsin(−2cos(2x−z)sin(x+z))+2πn−2x−2z,y=π+arcsin(2cos(2x−z)sin(x+z))+2πn−2x−2z