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受欢迎的 三角函数 >

sin(θ)-0.2cos(θ)=0.704

  • 初等代数
  • 代数
  • 微积分入门
  • 微积分
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  • 化学

解答

sin(θ)−0.2cos(θ)=0.704

解答

θ=2.57704…+2πn,θ=0.95933…+2πn
+1
度数
θ=147.65379…∘+360∘n,θ=54.96607…∘+360∘n
求解步骤
sin(θ)−0.2cos(θ)=0.704
两边加上 0.2cos(θ)sin(θ)=0.704+0.2cos(θ)
两边进行平方sin2(θ)=(0.704+0.2cos(θ))2
两边减去 (0.704+0.2cos(θ))2sin2(θ)−0.495616−0.2816cos(θ)−0.04cos2(θ)=0
使用三角恒等式改写
−0.495616+sin2(θ)−0.04cos2(θ)−0.2816cos(θ)
使用毕达哥拉斯恒等式: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−0.495616+1−cos2(θ)−0.04cos2(θ)−0.2816cos(θ)
化简 −0.495616+1−cos2(θ)−0.04cos2(θ)−0.2816cos(θ):−1.04cos2(θ)−0.2816cos(θ)+0.504384
−0.495616+1−cos2(θ)−0.04cos2(θ)−0.2816cos(θ)
同类项相加:−cos2(θ)−0.04cos2(θ)=−1.04cos2(θ)=−0.495616+1−1.04cos2(θ)−0.2816cos(θ)
数字相加/相减:−0.495616+1=0.504384=−1.04cos2(θ)−0.2816cos(θ)+0.504384
=−1.04cos2(θ)−0.2816cos(θ)+0.504384
0.504384−0.2816cos(θ)−1.04cos2(θ)=0
用替代法求解
0.504384−0.2816cos(θ)−1.04cos2(θ)=0
令:cos(θ)=u0.504384−0.2816u−1.04u2=0
0.504384−0.2816u−1.04u2=0:u=−2.080.2816+2.177536​​,u=2.082.177536​−0.2816​
0.504384−0.2816u−1.04u2=0
改写成标准形式 ax2+bx+c=0−1.04u2−0.2816u+0.504384=0
使用求根公式求解
−1.04u2−0.2816u+0.504384=0
二次方程求根公式:
若 a=−1.04,b=−0.2816,c=0.504384u1,2​=2(−1.04)−(−0.2816)±(−0.2816)2−4(−1.04)⋅0.504384​​
u1,2​=2(−1.04)−(−0.2816)±(−0.2816)2−4(−1.04)⋅0.504384​​
(−0.2816)2−4(−1.04)⋅0.504384​=2.177536​
(−0.2816)2−4(−1.04)⋅0.504384​
使用法则 −(−a)=a=(−0.2816)2+4⋅1.04⋅0.504384​
使用指数法则: (−a)n=an,若 n 是偶数(−0.2816)2=0.28162=0.28162+4⋅0.504384⋅1.04​
数字相乘:4⋅1.04⋅0.504384=2.09823744=0.28162+2.09823744​
0.28162=0.07929856=0.07929856+2.09823744​
数字相加:0.07929856+2.09823744=2.177536=2.177536​
u1,2​=2(−1.04)−(−0.2816)±2.177536​​
将解分隔开u1​=2(−1.04)−(−0.2816)+2.177536​​,u2​=2(−1.04)−(−0.2816)−2.177536​​
u=2(−1.04)−(−0.2816)+2.177536​​:−2.080.2816+2.177536​​
2(−1.04)−(−0.2816)+2.177536​​
去除括号: (−a)=−a,−(−a)=a=−2⋅1.040.2816+2.177536​​
数字相乘:2⋅1.04=2.08=−2.080.2816+2.177536​​
使用分式法则: −ba​=−ba​=−2.080.2816+2.177536​​
u=2(−1.04)−(−0.2816)−2.177536​​:2.082.177536​−0.2816​
2(−1.04)−(−0.2816)−2.177536​​
去除括号: (−a)=−a,−(−a)=a=−2⋅1.040.2816−2.177536​​
数字相乘:2⋅1.04=2.08=−2.080.2816−2.177536​​
使用分式法则: −b−a​=ba​0.2816−2.177536​=−(2.177536​−0.2816)=2.082.177536​−0.2816​
二次方程组的解是:u=−2.080.2816+2.177536​​,u=2.082.177536​−0.2816​
u=cos(θ)代回cos(θ)=−2.080.2816+2.177536​​,cos(θ)=2.082.177536​−0.2816​
cos(θ)=−2.080.2816+2.177536​​,cos(θ)=2.082.177536​−0.2816​
cos(θ)=−2.080.2816+2.177536​​:θ=arccos(−2.080.2816+2.177536​​)+2πn,θ=−arccos(−2.080.2816+2.177536​​)+2πn
cos(θ)=−2.080.2816+2.177536​​
使用反三角函数性质
cos(θ)=−2.080.2816+2.177536​​
cos(θ)=−2.080.2816+2.177536​​的通解cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnθ=arccos(−2.080.2816+2.177536​​)+2πn,θ=−arccos(−2.080.2816+2.177536​​)+2πn
θ=arccos(−2.080.2816+2.177536​​)+2πn,θ=−arccos(−2.080.2816+2.177536​​)+2πn
cos(θ)=2.082.177536​−0.2816​:θ=arccos(2.082.177536​−0.2816​)+2πn,θ=2π−arccos(2.082.177536​−0.2816​)+2πn
cos(θ)=2.082.177536​−0.2816​
使用反三角函数性质
cos(θ)=2.082.177536​−0.2816​
cos(θ)=2.082.177536​−0.2816​的通解cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnθ=arccos(2.082.177536​−0.2816​)+2πn,θ=2π−arccos(2.082.177536​−0.2816​)+2πn
θ=arccos(2.082.177536​−0.2816​)+2πn,θ=2π−arccos(2.082.177536​−0.2816​)+2πn
合并所有解θ=arccos(−2.080.2816+2.177536​​)+2πn,θ=−arccos(−2.080.2816+2.177536​​)+2πn,θ=arccos(2.082.177536​−0.2816​)+2πn,θ=2π−arccos(2.082.177536​−0.2816​)+2πn
将解代入原方程进行验证
将它们代入 sin(θ)−0.2cos(θ)=0.704检验解是否符合
去除与方程不符的解。
检验 arccos(−2.080.2816+2.177536​​)+2πn的解:真
arccos(−2.080.2816+2.177536​​)+2πn
代入 n=1arccos(−2.080.2816+2.177536​​)+2π1
对于 sin(θ)−0.2cos(θ)=0.704代入θ=arccos(−2.080.2816+2.177536​​)+2π1sin(arccos(−2.080.2816+2.177536​​)+2π1)−0.2cos(arccos(−2.080.2816+2.177536​​)+2π1)=0.704
整理后得0.704=0.704
⇒真
检验 −arccos(−2.080.2816+2.177536​​)+2πn的解:假
−arccos(−2.080.2816+2.177536​​)+2πn
代入 n=1−arccos(−2.080.2816+2.177536​​)+2π1
对于 sin(θ)−0.2cos(θ)=0.704代入θ=−arccos(−2.080.2816+2.177536​​)+2π1sin(−arccos(−2.080.2816+2.177536​​)+2π1)−0.2cos(−arccos(−2.080.2816+2.177536​​)+2π1)=0.704
整理后得−0.36606…=0.704
⇒假
检验 arccos(2.082.177536​−0.2816​)+2πn的解:真
arccos(2.082.177536​−0.2816​)+2πn
代入 n=1arccos(2.082.177536​−0.2816​)+2π1
对于 sin(θ)−0.2cos(θ)=0.704代入θ=arccos(2.082.177536​−0.2816​)+2π1sin(arccos(2.082.177536​−0.2816​)+2π1)−0.2cos(arccos(2.082.177536​−0.2816​)+2π1)=0.704
整理后得0.704=0.704
⇒真
检验 2π−arccos(2.082.177536​−0.2816​)+2πn的解:假
2π−arccos(2.082.177536​−0.2816​)+2πn
代入 n=12π−arccos(2.082.177536​−0.2816​)+2π1
对于 sin(θ)−0.2cos(θ)=0.704代入θ=2π−arccos(2.082.177536​−0.2816​)+2π1sin(2π−arccos(2.082.177536​−0.2816​)+2π1)−0.2cos(2π−arccos(2.082.177536​−0.2816​)+2π1)=0.704
整理后得−0.93362…=0.704
⇒假
θ=arccos(−2.080.2816+2.177536​​)+2πn,θ=arccos(2.082.177536​−0.2816​)+2πn
以小数形式表示解θ=2.57704…+2πn,θ=0.95933…+2πn

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