解
sin(θ)=0.6367+0.1⋅cos(θ)
解
θ=2.55514…+2πn,θ=0.78578…+2πn
+1
度
θ=146.39879…∘+360∘n,θ=45.02238…∘+360∘n解答ステップ
sin(θ)=0.6367+0.1cos(θ)
両辺を2乗するsin2(θ)=(0.6367+0.1cos(θ))2
両辺から(0.6367+0.1cos(θ))2を引くsin2(θ)−0.40538689−0.12734cos(θ)−0.01cos2(θ)=0
三角関数の公式を使用して書き換える
−0.40538689+sin2(θ)−0.01cos2(θ)−0.12734cos(θ)
ピタゴラスの公式を使用する: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.40538689+1−cos2(θ)−0.01cos2(θ)−0.12734cos(θ)
簡素化 −0.40538689+1−cos2(θ)−0.01cos2(θ)−0.12734cos(θ):−1.01cos2(θ)−0.12734cos(θ)+0.59461311
−0.40538689+1−cos2(θ)−0.01cos2(θ)−0.12734cos(θ)
類似した元を足す:−cos2(θ)−0.01cos2(θ)=−1.01cos2(θ)=−0.40538689+1−1.01cos2(θ)−0.12734cos(θ)
数を足す/引く:−0.40538689+1=0.59461311=−1.01cos2(θ)−0.12734cos(θ)+0.59461311
=−1.01cos2(θ)−0.12734cos(θ)+0.59461311
0.59461311−0.12734cos(θ)−1.01cos2(θ)=0
置換で解く
0.59461311−0.12734cos(θ)−1.01cos2(θ)=0
仮定:cos(θ)=u0.59461311−0.12734u−1.01u2=0
0.59461311−0.12734u−1.01u2=0:u=−2.020.12734+2.41845244,u=2.022.41845244−0.12734
0.59461311−0.12734u−1.01u2=0
標準的な形式で書く ax2+bx+c=0−1.01u2−0.12734u+0.59461311=0
解くとthe二次式
−1.01u2−0.12734u+0.59461311=0
二次Equationの公式:
次の場合: a=−1.01,b=−0.12734,c=0.59461311u1,2=2(−1.01)−(−0.12734)±(−0.12734)2−4(−1.01)⋅0.59461311
u1,2=2(−1.01)−(−0.12734)±(−0.12734)2−4(−1.01)⋅0.59461311
(−0.12734)2−4(−1.01)⋅0.59461311=2.41845244
(−0.12734)2−4(−1.01)⋅0.59461311
規則を適用 −(−a)=a=(−0.12734)2+4⋅1.01⋅0.59461311
指数の規則を適用する: n が偶数であれば (−a)n=an(−0.12734)2=0.127342=0.127342+4⋅0.59461311⋅1.01
数を乗じる:4⋅1.01⋅0.59461311=2.40223…=0.127342+2.40223…
0.127342=0.0162154756=0.0162154756+2.40223…
数を足す:0.0162154756+2.40223…=2.41845244=2.41845244
u1,2=2(−1.01)−(−0.12734)±2.41845244
解を分離するu1=2(−1.01)−(−0.12734)+2.41845244,u2=2(−1.01)−(−0.12734)−2.41845244
u=2(−1.01)−(−0.12734)+2.41845244:−2.020.12734+2.41845244
2(−1.01)−(−0.12734)+2.41845244
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅1.010.12734+2.41845244
数を乗じる:2⋅1.01=2.02=−2.020.12734+2.41845244
分数の規則を適用する: −ba=−ba=−2.020.12734+2.41845244
u=2(−1.01)−(−0.12734)−2.41845244:2.022.41845244−0.12734
2(−1.01)−(−0.12734)−2.41845244
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅1.010.12734−2.41845244
数を乗じる:2⋅1.01=2.02=−2.020.12734−2.41845244
分数の規則を適用する: −b−a=ba0.12734−2.41845244=−(2.41845244−0.12734)=2.022.41845244−0.12734
二次equationの解:u=−2.020.12734+2.41845244,u=2.022.41845244−0.12734
代用を戻す u=cos(θ)cos(θ)=−2.020.12734+2.41845244,cos(θ)=2.022.41845244−0.12734
cos(θ)=−2.020.12734+2.41845244,cos(θ)=2.022.41845244−0.12734
cos(θ)=−2.020.12734+2.41845244:θ=arccos(−2.020.12734+2.41845244)+2πn,θ=−arccos(−2.020.12734+2.41845244)+2πn
cos(θ)=−2.020.12734+2.41845244
三角関数の逆数プロパティを適用する
cos(θ)=−2.020.12734+2.41845244
以下の一般解 cos(θ)=−2.020.12734+2.41845244cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.020.12734+2.41845244)+2πn,θ=−arccos(−2.020.12734+2.41845244)+2πn
θ=arccos(−2.020.12734+2.41845244)+2πn,θ=−arccos(−2.020.12734+2.41845244)+2πn
cos(θ)=2.022.41845244−0.12734:θ=arccos(2.022.41845244−0.12734)+2πn,θ=2π−arccos(2.022.41845244−0.12734)+2πn
cos(θ)=2.022.41845244−0.12734
三角関数の逆数プロパティを適用する
cos(θ)=2.022.41845244−0.12734
以下の一般解 cos(θ)=2.022.41845244−0.12734cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.022.41845244−0.12734)+2πn,θ=2π−arccos(2.022.41845244−0.12734)+2πn
θ=arccos(2.022.41845244−0.12734)+2πn,θ=2π−arccos(2.022.41845244−0.12734)+2πn
すべての解を組み合わせるθ=arccos(−2.020.12734+2.41845244)+2πn,θ=−arccos(−2.020.12734+2.41845244)+2πn,θ=arccos(2.022.41845244−0.12734)+2πn,θ=2π−arccos(2.022.41845244−0.12734)+2πn
元のequationに当てはめて解を検算する
sin(θ)=0.6367+0.1cos(θ) に当てはめて解を確認する
equationに一致しないものを削除する。
解答を確認する arccos(−2.020.12734+2.41845244)+2πn:真
arccos(−2.020.12734+2.41845244)+2πn
挿入 n=1arccos(−2.020.12734+2.41845244)+2π1
sin(θ)=0.6367+0.1cos(θ)の挿入向けθ=arccos(−2.020.12734+2.41845244)+2π1sin(arccos(−2.020.12734+2.41845244)+2π1)=0.6367+0.1cos(arccos(−2.020.12734+2.41845244)+2π1)
改良0.55340…=0.55340…
⇒真
解答を確認する −arccos(−2.020.12734+2.41845244)+2πn:偽
−arccos(−2.020.12734+2.41845244)+2πn
挿入 n=1−arccos(−2.020.12734+2.41845244)+2π1
sin(θ)=0.6367+0.1cos(θ)の挿入向けθ=−arccos(−2.020.12734+2.41845244)+2π1sin(−arccos(−2.020.12734+2.41845244)+2π1)=0.6367+0.1cos(−arccos(−2.020.12734+2.41845244)+2π1)
改良−0.55340…=0.55340…
⇒偽
解答を確認する arccos(2.022.41845244−0.12734)+2πn:真
arccos(2.022.41845244−0.12734)+2πn
挿入 n=1arccos(2.022.41845244−0.12734)+2π1
sin(θ)=0.6367+0.1cos(θ)の挿入向けθ=arccos(2.022.41845244−0.12734)+2π1sin(arccos(2.022.41845244−0.12734)+2π1)=0.6367+0.1cos(arccos(2.022.41845244−0.12734)+2π1)
改良0.70738…=0.70738…
⇒真
解答を確認する 2π−arccos(2.022.41845244−0.12734)+2πn:偽
2π−arccos(2.022.41845244−0.12734)+2πn
挿入 n=12π−arccos(2.022.41845244−0.12734)+2π1
sin(θ)=0.6367+0.1cos(θ)の挿入向けθ=2π−arccos(2.022.41845244−0.12734)+2π1sin(2π−arccos(2.022.41845244−0.12734)+2π1)=0.6367+0.1cos(2π−arccos(2.022.41845244−0.12734)+2π1)
改良−0.70738…=0.70738…
⇒偽
θ=arccos(−2.020.12734+2.41845244)+2πn,θ=arccos(2.022.41845244−0.12734)+2πn
10進法形式で解を証明するθ=2.55514…+2πn,θ=0.78578…+2πn