解
sin(θ)−0.2cos(θ)=9.86.25
解
θ=2.66338…+2πn,θ=0.87300…+2πn
+1
度
θ=152.60055…∘+360∘n,θ=50.01930…∘+360∘n解答ステップ
sin(θ)−0.2cos(θ)=9.86.25
両辺に0.2cos(θ)を足すsin(θ)=0.63775…+0.2cos(θ)
両辺を2乗するsin2(θ)=(0.63775…+0.2cos(θ))2
両辺から(0.63775…+0.2cos(θ))2を引くsin2(θ)−0.40673…−0.25510…cos(θ)−0.04cos2(θ)=0
三角関数の公式を使用して書き換える
−0.40673…+sin2(θ)−0.04cos2(θ)−0.25510…cos(θ)
ピタゴラスの公式を使用する: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ)
簡素化 −0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ):−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
−0.40673…+1−cos2(θ)−0.04cos2(θ)−0.25510…cos(θ)
類似した元を足す:−cos2(θ)−0.04cos2(θ)=−1.04cos2(θ)=−0.40673…+1−1.04cos2(θ)−0.25510…cos(θ)
数を足す/引く:−0.40673…+1=0.59326…=−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
=−1.04cos2(θ)−0.25510…cos(θ)+0.59326…
0.59326…−0.25510…cos(θ)−1.04cos2(θ)=0
置換で解く
0.59326…−0.25510…cos(θ)−1.04cos2(θ)=0
仮定:cos(θ)=u0.59326…−0.25510…u−1.04u2=0
0.59326…−0.25510…u−1.04u2=0:u=−2.080.25510…+2.53307…,u=2.082.53307…−0.25510…
0.59326…−0.25510…u−1.04u2=0
標準的な形式で書く ax2+bx+c=0−1.04u2−0.25510…u+0.59326…=0
解くとthe二次式
−1.04u2−0.25510…u+0.59326…=0
二次Equationの公式:
次の場合: a=−1.04,b=−0.25510…,c=0.59326…u1,2=2(−1.04)−(−0.25510…)±(−0.25510…)2−4(−1.04)⋅0.59326…
u1,2=2(−1.04)−(−0.25510…)±(−0.25510…)2−4(−1.04)⋅0.59326…
(−0.25510…)2−4(−1.04)⋅0.59326…=2.53307…
(−0.25510…)2−4(−1.04)⋅0.59326…
規則を適用 −(−a)=a=(−0.25510…)2+4⋅1.04⋅0.59326…
指数の規則を適用する: n が偶数であれば (−a)n=an(−0.25510…)2=0.25510…2=0.25510…2+4⋅0.59326…⋅1.04
数を乗じる:4⋅1.04⋅0.59326…=2.46799…=0.25510…2+2.46799…
0.25510…2=0.06507…=0.06507…+2.46799…
数を足す:0.06507…+2.46799…=2.53307…=2.53307…
u1,2=2(−1.04)−(−0.25510…)±2.53307…
解を分離するu1=2(−1.04)−(−0.25510…)+2.53307…,u2=2(−1.04)−(−0.25510…)−2.53307…
u=2(−1.04)−(−0.25510…)+2.53307…:−2.080.25510…+2.53307…
2(−1.04)−(−0.25510…)+2.53307…
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅1.040.25510…+2.53307…
数を乗じる:2⋅1.04=2.08=−2.080.25510…+2.53307…
分数の規則を適用する: −ba=−ba=−2.080.25510…+2.53307…
u=2(−1.04)−(−0.25510…)−2.53307…:2.082.53307…−0.25510…
2(−1.04)−(−0.25510…)−2.53307…
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅1.040.25510…−2.53307…
数を乗じる:2⋅1.04=2.08=−2.080.25510…−2.53307…
分数の規則を適用する: −b−a=ba0.25510…−2.53307…=−(2.53307…−0.25510…)=2.082.53307…−0.25510…
二次equationの解:u=−2.080.25510…+2.53307…,u=2.082.53307…−0.25510…
代用を戻す u=cos(θ)cos(θ)=−2.080.25510…+2.53307…,cos(θ)=2.082.53307…−0.25510…
cos(θ)=−2.080.25510…+2.53307…,cos(θ)=2.082.53307…−0.25510…
cos(θ)=−2.080.25510…+2.53307…:θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
cos(θ)=−2.080.25510…+2.53307…
三角関数の逆数プロパティを適用する
cos(θ)=−2.080.25510…+2.53307…
以下の一般解 cos(θ)=−2.080.25510…+2.53307…cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn
cos(θ)=2.082.53307…−0.25510…:θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
cos(θ)=2.082.53307…−0.25510…
三角関数の逆数プロパティを適用する
cos(θ)=2.082.53307…−0.25510…
以下の一般解 cos(θ)=2.082.53307…−0.25510…cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
すべての解を組み合わせるθ=arccos(−2.080.25510…+2.53307…)+2πn,θ=−arccos(−2.080.25510…+2.53307…)+2πn,θ=arccos(2.082.53307…−0.25510…)+2πn,θ=2π−arccos(2.082.53307…−0.25510…)+2πn
元のequationに当てはめて解を検算する
sin(θ)−0.2cos(θ)=9.86.25 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(−2.080.25510…+2.53307…)+2πn:真
arccos(−2.080.25510…+2.53307…)+2πn
挿入 n=1arccos(−2.080.25510…+2.53307…)+2π1
sin(θ)−0.2cos(θ)=9.86.25の挿入向けθ=arccos(−2.080.25510…+2.53307…)+2π1sin(arccos(−2.080.25510…+2.53307…)+2π1)−0.2cos(arccos(−2.080.25510…+2.53307…)+2π1)=9.86.25
改良0.63775…=0.63775…
⇒真
解答を確認する −arccos(−2.080.25510…+2.53307…)+2πn:偽
−arccos(−2.080.25510…+2.53307…)+2πn
挿入 n=1−arccos(−2.080.25510…+2.53307…)+2π1
sin(θ)−0.2cos(θ)=9.86.25の挿入向けθ=−arccos(−2.080.25510…+2.53307…)+2π1sin(−arccos(−2.080.25510…+2.53307…)+2π1)−0.2cos(−arccos(−2.080.25510…+2.53307…)+2π1)=9.86.25
改良−0.28262…=0.63775…
⇒偽
解答を確認する arccos(2.082.53307…−0.25510…)+2πn:真
arccos(2.082.53307…−0.25510…)+2πn
挿入 n=1arccos(2.082.53307…−0.25510…)+2π1
sin(θ)−0.2cos(θ)=9.86.25の挿入向けθ=arccos(2.082.53307…−0.25510…)+2π1sin(arccos(2.082.53307…−0.25510…)+2π1)−0.2cos(arccos(2.082.53307…−0.25510…)+2π1)=9.86.25
改良0.63775…=0.63775…
⇒真
解答を確認する 2π−arccos(2.082.53307…−0.25510…)+2πn:偽
2π−arccos(2.082.53307…−0.25510…)+2πn
挿入 n=12π−arccos(2.082.53307…−0.25510…)+2π1
sin(θ)−0.2cos(θ)=9.86.25の挿入向けθ=2π−arccos(2.082.53307…−0.25510…)+2π1sin(2π−arccos(2.082.53307…−0.25510…)+2π1)−0.2cos(2π−arccos(2.082.53307…−0.25510…)+2π1)=9.86.25
改良−0.89476…=0.63775…
⇒偽
θ=arccos(−2.080.25510…+2.53307…)+2πn,θ=arccos(2.082.53307…−0.25510…)+2πn
10進法形式で解を証明するθ=2.66338…+2πn,θ=0.87300…+2πn