解
49.55cos(θ)−30sin(θ)=1.225
解
θ=π+1.04752…+2πn,θ=1.00522…+2πn
+1
度
θ=240.01903…∘+360∘n,θ=57.59542…∘+360∘n解答ステップ
49.55cos(θ)−30sin(θ)=1.225
両辺に30sin(θ)を足す49.55cos(θ)=1.225+30sin(θ)
両辺を2乗する(49.55cos(θ))2=(1.225+30sin(θ))2
両辺から(1.225+30sin(θ))2を引く2455.2025cos2(θ)−1.500625−73.5sin(θ)−900sin2(θ)=0
三角関数の公式を使用して書き換える
−1.500625+2455.2025cos2(θ)−73.5sin(θ)−900sin2(θ)
ピタゴラスの公式を使用する: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=−1.500625+2455.2025(1−sin2(θ))−73.5sin(θ)−900sin2(θ)
簡素化 −1.500625+2455.2025(1−sin2(θ))−73.5sin(θ)−900sin2(θ):−3355.2025sin2(θ)−73.5sin(θ)+2453.701875
−1.500625+2455.2025(1−sin2(θ))−73.5sin(θ)−900sin2(θ)
拡張 2455.2025(1−sin2(θ)):2455.2025−2455.2025sin2(θ)
2455.2025(1−sin2(θ))
分配法則を適用する: a(b−c)=ab−aca=2455.2025,b=1,c=sin2(θ)=2455.2025⋅1−2455.2025sin2(θ)
=1⋅2455.2025−2455.2025sin2(θ)
数を乗じる:1⋅2455.2025=2455.2025=2455.2025−2455.2025sin2(θ)
=−1.500625+2455.2025−2455.2025sin2(θ)−73.5sin(θ)−900sin2(θ)
簡素化 −1.500625+2455.2025−2455.2025sin2(θ)−73.5sin(θ)−900sin2(θ):−3355.2025sin2(θ)−73.5sin(θ)+2453.701875
−1.500625+2455.2025−2455.2025sin2(θ)−73.5sin(θ)−900sin2(θ)
条件のようなグループ=−2455.2025sin2(θ)−73.5sin(θ)−900sin2(θ)−1.500625+2455.2025
類似した元を足す:−2455.2025sin2(θ)−900sin2(θ)=−3355.2025sin2(θ)=−3355.2025sin2(θ)−73.5sin(θ)−1.500625+2455.2025
数を足す/引く:−1.500625+2455.2025=2453.701875=−3355.2025sin2(θ)−73.5sin(θ)+2453.701875
=−3355.2025sin2(θ)−73.5sin(θ)+2453.701875
=−3355.2025sin2(θ)−73.5sin(θ)+2453.701875
2453.701875−3355.2025sin2(θ)−73.5sin(θ)=0
置換で解く
2453.701875−3355.2025sin2(θ)−73.5sin(θ)=0
仮定:sin(θ)=u2453.701875−3355.2025u2−73.5u=0
2453.701875−3355.2025u2−73.5u=0:u=−6710.40573.5+32936068.91101…,u=6710.40532936068.91101…−73.5
2453.701875−3355.2025u2−73.5u=0
標準的な形式で書く ax2+bx+c=0−3355.2025u2−73.5u+2453.701875=0
解くとthe二次式
−3355.2025u2−73.5u+2453.701875=0
二次Equationの公式:
次の場合: a=−3355.2025,b=−73.5,c=2453.701875u1,2=2(−3355.2025)−(−73.5)±(−73.5)2−4(−3355.2025)⋅2453.701875
u1,2=2(−3355.2025)−(−73.5)±(−73.5)2−4(−3355.2025)⋅2453.701875
(−73.5)2−4(−3355.2025)⋅2453.701875=32936068.91101…
(−73.5)2−4(−3355.2025)⋅2453.701875
規則を適用 −(−a)=a=(−73.5)2+4⋅3355.2025⋅2453.701875
指数の規則を適用する: n が偶数であれば (−a)n=an(−73.5)2=73.52=73.52+4⋅2453.701875⋅3355.2025
数を乗じる:4⋅3355.2025⋅2453.701875=32930666.66101…=73.52+32930666.66101…
73.52=5402.25=5402.25+32930666.66101…
数を足す:5402.25+32930666.66101…=32936068.91101…=32936068.91101…
u1,2=2(−3355.2025)−(−73.5)±32936068.91101…
解を分離するu1=2(−3355.2025)−(−73.5)+32936068.91101…,u2=2(−3355.2025)−(−73.5)−32936068.91101…
u=2(−3355.2025)−(−73.5)+32936068.91101…:−6710.40573.5+32936068.91101…
2(−3355.2025)−(−73.5)+32936068.91101…
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅3355.202573.5+32936068.91101…
数を乗じる:2⋅3355.2025=6710.405=−6710.40573.5+32936068.91101…
分数の規則を適用する: −ba=−ba=−6710.40573.5+32936068.91101…
u=2(−3355.2025)−(−73.5)−32936068.91101…:6710.40532936068.91101…−73.5
2(−3355.2025)−(−73.5)−32936068.91101…
括弧を削除する: (−a)=−a,−(−a)=a=−2⋅3355.202573.5−32936068.91101…
数を乗じる:2⋅3355.2025=6710.405=−6710.40573.5−32936068.91101…
分数の規則を適用する: −b−a=ba73.5−32936068.91101…=−(32936068.91101…−73.5)=6710.40532936068.91101…−73.5
二次equationの解:u=−6710.40573.5+32936068.91101…,u=6710.40532936068.91101…−73.5
代用を戻す u=sin(θ)sin(θ)=−6710.40573.5+32936068.91101…,sin(θ)=6710.40532936068.91101…−73.5
sin(θ)=−6710.40573.5+32936068.91101…,sin(θ)=6710.40532936068.91101…−73.5
sin(θ)=−6710.40573.5+32936068.91101…:θ=arcsin(−6710.40573.5+32936068.91101…)+2πn,θ=π+arcsin(6710.40573.5+32936068.91101…)+2πn
sin(θ)=−6710.40573.5+32936068.91101…
三角関数の逆数プロパティを適用する
sin(θ)=−6710.40573.5+32936068.91101…
以下の一般解 sin(θ)=−6710.40573.5+32936068.91101…sin(x)=−a⇒x=arcsin(−a)+2πn,x=π+arcsin(a)+2πnθ=arcsin(−6710.40573.5+32936068.91101…)+2πn,θ=π+arcsin(6710.40573.5+32936068.91101…)+2πn
θ=arcsin(−6710.40573.5+32936068.91101…)+2πn,θ=π+arcsin(6710.40573.5+32936068.91101…)+2πn
sin(θ)=6710.40532936068.91101…−73.5:θ=arcsin(6710.40532936068.91101…−73.5)+2πn,θ=π−arcsin(6710.40532936068.91101…−73.5)+2πn
sin(θ)=6710.40532936068.91101…−73.5
三角関数の逆数プロパティを適用する
sin(θ)=6710.40532936068.91101…−73.5
以下の一般解 sin(θ)=6710.40532936068.91101…−73.5sin(x)=a⇒x=arcsin(a)+2πn,x=π−arcsin(a)+2πnθ=arcsin(6710.40532936068.91101…−73.5)+2πn,θ=π−arcsin(6710.40532936068.91101…−73.5)+2πn
θ=arcsin(6710.40532936068.91101…−73.5)+2πn,θ=π−arcsin(6710.40532936068.91101…−73.5)+2πn
すべての解を組み合わせるθ=arcsin(−6710.40573.5+32936068.91101…)+2πn,θ=π+arcsin(6710.40573.5+32936068.91101…)+2πn,θ=arcsin(6710.40532936068.91101…−73.5)+2πn,θ=π−arcsin(6710.40532936068.91101…−73.5)+2πn
元のequationに当てはめて解を検算する
49.55cos(θ)−30sin(θ)=1.225 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 arcsin(−6710.40573.5+32936068.91101…)+2πn:偽
arcsin(−6710.40573.5+32936068.91101…)+2πn
挿入 n=1arcsin(−6710.40573.5+32936068.91101…)+2π1
49.55cos(θ)−30sin(θ)=1.225の挿入向けθ=arcsin(−6710.40573.5+32936068.91101…)+2π149.55cos(arcsin(−6710.40573.5+32936068.91101…)+2π1)−30sin(arcsin(−6710.40573.5+32936068.91101…)+2π1)=1.225
改良50.74648…=1.225
⇒偽
解答を確認する π+arcsin(6710.40573.5+32936068.91101…)+2πn:真
π+arcsin(6710.40573.5+32936068.91101…)+2πn
挿入 n=1π+arcsin(6710.40573.5+32936068.91101…)+2π1
49.55cos(θ)−30sin(θ)=1.225の挿入向けθ=π+arcsin(6710.40573.5+32936068.91101…)+2π149.55cos(π+arcsin(6710.40573.5+32936068.91101…)+2π1)−30sin(π+arcsin(6710.40573.5+32936068.91101…)+2π1)=1.225
改良1.22499…=1.225
⇒真
解答を確認する arcsin(6710.40532936068.91101…−73.5)+2πn:真
arcsin(6710.40532936068.91101…−73.5)+2πn
挿入 n=1arcsin(6710.40532936068.91101…−73.5)+2π1
49.55cos(θ)−30sin(θ)=1.225の挿入向けθ=arcsin(6710.40532936068.91101…−73.5)+2π149.55cos(arcsin(6710.40532936068.91101…−73.5)+2π1)−30sin(arcsin(6710.40532936068.91101…−73.5)+2π1)=1.225
改良1.225=1.225
⇒真
解答を確認する π−arcsin(6710.40532936068.91101…−73.5)+2πn:偽
π−arcsin(6710.40532936068.91101…−73.5)+2πn
挿入 n=1π−arcsin(6710.40532936068.91101…−73.5)+2π1
49.55cos(θ)−30sin(θ)=1.225の挿入向けθ=π−arcsin(6710.40532936068.91101…−73.5)+2π149.55cos(π−arcsin(6710.40532936068.91101…−73.5)+2π1)−30sin(π−arcsin(6710.40532936068.91101…−73.5)+2π1)=1.225
改良−51.88211…=1.225
⇒偽
θ=π+arcsin(6710.40573.5+32936068.91101…)+2πn,θ=arcsin(6710.40532936068.91101…−73.5)+2πn
10進法形式で解を証明するθ=π+1.04752…+2πn,θ=1.00522…+2πn