解答
cos5(x)−sin(x)=0
解答
x=0.51794…+2πn,x=−2.62364…+2πn
+1
度数
x=29.67623…∘+360∘n,x=−150.32376…∘+360∘n求解步骤
cos5(x)−sin(x)=0
两边加上 sin(x)cos5(x)=sin(x)
两边进行平方(cos5(x))2=sin2(x)
两边减去 sin2(x)cos10(x)−sin2(x)=0
使用三角恒等式改写
cos10(x)−sin2(x)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=cos10(x)−(1−cos2(x))
−(1−cos2(x)):−1+cos2(x)
−(1−cos2(x))
打开括号=−(1)−(−cos2(x))
使用加减运算法则−(−a)=a,−(a)=−a=−1+cos2(x)
=cos10(x)−1+cos2(x)
−1+cos10(x)+cos2(x)=0
用替代法求解
−1+cos10(x)+cos2(x)=0
令:cos(x)=u−1+u10+u2=0
−1+u10+u2=0:u=0.75487…,u=−0.75487…
−1+u10+u2=0
改写成标准形式 anxn+…+a1x+a0=0u10+u2−1=0
用v=u2 和 v5=u10改写方程式v5+v−1=0
解 v5+v−1=0:v≈0.75487…
v5+v−1=0
使用牛顿-拉弗森方法找到 v5+v−1=0 的一个解:v≈0.75487…
v5+v−1=0
牛顿-拉弗森近似法定义
f(v)=v5+v−1
找到 f′(v):5v4+1
dvd(v5+v−1)
使用微分加减法定则: (f±g)′=f′±g′=dvd(v5)+dvdv−dvd(1)
dvd(v5)=5v4
dvd(v5)
使用幂法则: dxd(xa)=a⋅xa−1=5v5−1
化简=5v4
dvdv=1
dvdv
使用常见微分定则: dvdv=1=1
dvd(1)=0
dvd(1)
常数微分: dxd(a)=0=0
=5v4+1−0
化简=5v4+1
令 v0=1计算 vn+1 至 Δvn+1<0.000001
v1=0.83333…:Δv1=0.16666…
f(v0)=15+1−1=1f′(v0)=5⋅14+1=6v1=0.83333…
Δv1=∣0.83333…−1∣=0.16666…Δv1=0.16666…
v2=0.76438…:Δv2=0.06895…
f(v1)=0.83333…5+0.83333…−1=0.23521…f′(v1)=5⋅0.83333…4+1=3.41126…v2=0.76438…
Δv2=∣0.76438…−0.83333…∣=0.06895…Δv2=0.06895…
v3=0.75502…:Δv3=0.00935…
f(v2)=0.76438…5+0.76438…−1=0.02532…f′(v2)=5⋅0.76438…4+1=2.70691…v3=0.75502…
Δv3=∣0.75502…−0.76438…∣=0.00935…Δv3=0.00935…
v4=0.75487…:Δv4=0.00014…
f(v3)=0.75502…5+0.75502…−1=0.00038…f′(v3)=5⋅0.75502…4+1=2.62485…v4=0.75487…
Δv4=∣0.75487…−0.75502…∣=0.00014…Δv4=0.00014…
v5=0.75487…:Δv5=3.55234E−8
f(v4)=0.75487…5+0.75487…−1=9.31989E−8f′(v4)=5⋅0.75487…4+1=2.62359…v5=0.75487…
Δv5=∣0.75487…−0.75487…∣=3.55234E−8Δv5=3.55234E−8
v≈0.75487…
使用长除法 Equation0:v−0.75487…v5+v−1=v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…
v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…≈0
使用牛顿-拉弗森方法找到 v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…=0 的一个解:v∈R无解
v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…=0
牛顿-拉弗森近似法定义
f(v)=v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…
找到 f′(v):4v3+2.26463…v2+1.13968…v+0.43015…
dvd(v4+0.75487…v3+0.56984…v2+0.43015…v+1.32471…)
使用微分加减法定则: (f±g)′=f′±g′=dvd(v4)+dvd(0.75487…v3)+dvd(0.56984…v2)+dvd(0.43015…v)+dvd(1.32471…)
dvd(v4)=4v3
dvd(v4)
使用幂法则: dxd(xa)=a⋅xa−1=4v4−1
化简=4v3
dvd(0.75487…v3)=2.26463…v2
dvd(0.75487…v3)
将常数提出: (a⋅f)′=a⋅f′=0.75487…dvd(v3)
使用幂法则: dxd(xa)=a⋅xa−1=0.75487…⋅3v3−1
化简=2.26463…v2
dvd(0.56984…v2)=1.13968…v
dvd(0.56984…v2)
将常数提出: (a⋅f)′=a⋅f′=0.56984…dvd(v2)
使用幂法则: dxd(xa)=a⋅xa−1=0.56984…⋅2v2−1
化简=1.13968…v
dvd(0.43015…v)=0.43015…
dvd(0.43015…v)
将常数提出: (a⋅f)′=a⋅f′=0.43015…dvdv
使用常见微分定则: dvdv=1=0.43015…⋅1
化简=0.43015…
dvd(1.32471…)=0
dvd(1.32471…)
常数微分: dxd(a)=0=0
=4v3+2.26463…v2+1.13968…v+0.43015…+0
化简=4v3+2.26463…v2+1.13968…v+0.43015…
令 v0=−3计算 vn+1 至 Δvn+1<0.000001
v1=−2.27399…:Δv1=0.72600…
f(v0)=(−3)4+0.75487…(−3)3+0.56984…(−3)2+0.43015…(−3)+1.32471…=65.78110…f′(v0)=4(−3)3+2.26463…(−3)2+1.13968…(−3)+0.43015…=−90.60718…v1=−2.27399…
Δv1=∣−2.27399…−(−3)∣=0.72600…Δv1=0.72600…
v2=−1.70962…:Δv2=0.56437…
f(v1)=(−2.27399…)4+0.75487…(−2.27399…)3+0.56984…(−2.27399…)2+0.43015…(−2.27399…)+1.32471…=21.15650…f′(v1)=4(−2.27399…)3+2.26463…(−2.27399…)2+1.13968…(−2.27399…)+0.43015…=−37.48682…v2=−1.70962…
Δv2=∣−1.70962…−(−2.27399…)∣=0.56437…Δv2=0.56437…
v3=−1.23768…:Δv3=0.47193…
f(v2)=(−1.70962…)4+0.75487…(−1.70962…)3+0.56984…(−1.70962…)2+0.43015…(−1.70962…)+1.32471…=7.02564…f′(v2)=4(−1.70962…)3+2.26463…(−1.70962…)2+1.13968…(−1.70962…)+0.43015…=−14.88684…v3=−1.23768…
Δv3=∣−1.23768…−(−1.70962…)∣=0.47193…Δv3=0.47193…
v4=−0.73120…:Δv4=0.50648…
f(v3)=(−1.23768…)4+0.75487…(−1.23768…)3+0.56984…(−1.23768…)2+0.43015…(−1.23768…)+1.32471…=2.58063…f′(v3)=4(−1.23768…)3+2.26463…(−1.23768…)2+1.13968…(−1.23768…)+0.43015…=−5.09520…v4=−0.73120…
Δv4=∣−0.73120…−(−1.23768…)∣=0.50648…Δv4=0.50648…
v5=0.99541…:Δv5=1.72662…
f(v4)=(−0.73120…)4+0.75487…(−0.73120…)3+0.56984…(−0.73120…)2+0.43015…(−0.73120…)+1.32471…=1.30559…f′(v4)=4(−0.73120…)3+2.26463…(−0.73120…)2+1.13968…(−0.73120…)+0.43015…=−0.75615…v5=0.99541…
Δv5=∣0.99541…−(−0.73120…)∣=1.72662…Δv5=1.72662…
v6=0.47388…:Δv6=0.52153…
f(v5)=0.99541…4+0.75487…⋅0.99541…3+0.56984…⋅0.99541…2+0.43015…⋅0.99541…+1.32471…=4.04387…f′(v5)=4⋅0.99541…3+2.26463…⋅0.99541…2+1.13968…⋅0.99541…+0.43015…=7.75380…v6=0.47388…
Δv6=∣0.47388…−0.99541…∣=0.52153…Δv6=0.52153…
v7=−0.46459…:Δv7=0.93847…
f(v6)=0.47388…4+0.75487…⋅0.47388…3+0.56984…⋅0.47388…2+0.43015…⋅0.47388…+1.32471…=1.78729…f′(v6)=4⋅0.47388…3+2.26463…⋅0.47388…2+1.13968…⋅0.47388…+0.43015…=1.90446…v7=−0.46459…
Δv7=∣−0.46459…−0.47388…∣=0.93847…Δv7=0.93847…
v8=104.25021…:Δv8=104.71480…
f(v7)=(−0.46459…)4+0.75487…(−0.46459…)3+0.56984…(−0.46459…)2+0.43015…(−0.46459…)+1.32471…=1.21875…f′(v7)=4(−0.46459…)3+2.26463…(−0.46459…)2+1.13968…(−0.46459…)+0.43015…=−0.01163…v8=104.25021…
Δv8=∣104.25021…−(−0.46459…)∣=104.71480…Δv8=104.71480…
无法得出解
解是v≈0.75487…
v≈0.75487…
代回 v=u2,求解 u
解 u2=0.75487…:u=0.75487…,u=−0.75487…
u2=0.75487…
对于 x2=f(a) 解为 x=f(a),−f(a)
u=0.75487…,u=−0.75487…
解为
u=0.75487…,u=−0.75487…
u=cos(x)代回cos(x)=0.75487…,cos(x)=−0.75487…
cos(x)=0.75487…,cos(x)=−0.75487…
cos(x)=0.75487…:x=arccos(0.75487…)+2πn,x=2π−arccos(0.75487…)+2πn
cos(x)=0.75487…
使用反三角函数性质
cos(x)=0.75487…
cos(x)=0.75487…的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(0.75487…)+2πn,x=2π−arccos(0.75487…)+2πn
x=arccos(0.75487…)+2πn,x=2π−arccos(0.75487…)+2πn
cos(x)=−0.75487…:x=arccos(−0.75487…)+2πn,x=−arccos(−0.75487…)+2πn
cos(x)=−0.75487…
使用反三角函数性质
cos(x)=−0.75487…
cos(x)=−0.75487…的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−0.75487…)+2πn,x=−arccos(−0.75487…)+2πn
x=arccos(−0.75487…)+2πn,x=−arccos(−0.75487…)+2πn
合并所有解x=arccos(0.75487…)+2πn,x=2π−arccos(0.75487…)+2πn,x=arccos(−0.75487…)+2πn,x=−arccos(−0.75487…)+2πn
将解代入原方程进行验证
将它们代入 cos5(x)−sin(x)=0检验解是否符合
去除与方程不符的解。
检验 arccos(0.75487…)+2πn的解:真
arccos(0.75487…)+2πn
代入 n=1arccos(0.75487…)+2π1
对于 cos5(x)−sin(x)=0代入x=arccos(0.75487…)+2π1cos5(arccos(0.75487…)+2π1)−sin(arccos(0.75487…)+2π1)=0
整理后得0=0
⇒真
检验 2π−arccos(0.75487…)+2πn的解:假
2π−arccos(0.75487…)+2πn
代入 n=12π−arccos(0.75487…)+2π1
对于 cos5(x)−sin(x)=0代入x=2π−arccos(0.75487…)+2π1cos5(2π−arccos(0.75487…)+2π1)−sin(2π−arccos(0.75487…)+2π1)=0
整理后得0.99019…=0
⇒假
检验 arccos(−0.75487…)+2πn的解:假
arccos(−0.75487…)+2πn
代入 n=1arccos(−0.75487…)+2π1
对于 cos5(x)−sin(x)=0代入x=arccos(−0.75487…)+2π1cos5(arccos(−0.75487…)+2π1)−sin(arccos(−0.75487…)+2π1)=0
整理后得−0.99019…=0
⇒假
检验 −arccos(−0.75487…)+2πn的解:真
−arccos(−0.75487…)+2πn
代入 n=1−arccos(−0.75487…)+2π1
对于 cos5(x)−sin(x)=0代入x=−arccos(−0.75487…)+2π1cos5(−arccos(−0.75487…)+2π1)−sin(−arccos(−0.75487…)+2π1)=0
整理后得0=0
⇒真
x=arccos(0.75487…)+2πn,x=−arccos(−0.75487…)+2πn
以小数形式表示解x=0.51794…+2πn,x=−2.62364…+2πn