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受欢迎的三角函数 >

2sin^3(x/2)cos(x/2)=cos^2(x/2)sin(x)

解答

2 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) = cos^{2} (\frac{x}{2}) sin (x)

解答

x = π + 4 πn, x = 3 π + 4 πn, x = 4 πn, x = 2 π + 4 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

x = 18 0^{\circ} + 72 0^{\circ} n, x = 54 0^{\circ} + 72 0^{\circ} n, x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) = cos^{2} (\frac{x}{2}) sin (x)

两边减去

cos^{2} (\frac{x}{2}) sin (x)

2 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) - cos^{2} (\frac{x}{2}) sin (x) = 0

令

: u = \frac{x}{2}

2 sin^{3} (u) cos (u) - cos^{2} (u) sin (2 u) = 0

分解

2 sin^{3} (u) cos (u) - cos^{2} (u) sin (2 u) : cos (u) (2 sin^{3} (u) - cos (u) sin (2 u))

2 sin^{3} (u) cos (u) - cos^{2} (u) sin (2 u)

使用指数法则:

a^{b + c} = a^{b} a^{c}

cos^{2} (u) = cos (u) cos (u)

= 2 sin^{3} (u) cos (u) - cos (u) cos (u) sin (2 u)

因式分解出通项

cos (u)

= cos (u) (2 sin^{3} (u) - cos (u) sin (2 u))

cos (u) (2 sin^{3} (u) - cos (u) sin (2 u)) = 0

分别求解每个部分

cos (u) = 0 or 2 sin^{3} (u) - cos (u) sin (2 u) = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

2 sin^{3} (u) - cos (u) sin (2 u) = 0 : u = 2 πn, u = π + 2 πn, u = \frac{3 π}{4} + πn, u = \frac{π}{4} + πn

2 sin^{3} (u) - cos (u) sin (2 u) = 0

使用三角恒等式改写

2 sin^{3} (u) - cos (u) sin (2 u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin^{3} (u) - cos (u) \cdot 2 sin (u) cos (u)

cos (u) \cdot 2 sin (u) cos (u) = 2 cos^{2} (u) sin (u)

cos (u) \cdot 2 sin (u) cos (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (u) cos (u) = cos^{1 + 1} (u)

= 2 sin (u) cos^{1 + 1} (u)

数字相加:

1 + 1 = 2

= 2 sin (u) cos^{2} (u)

= 2 sin^{3} (u) - 2 cos^{2} (u) sin (u)

2 sin^{3} (u) - 2 cos^{2} (u) sin (u) = 0

分解

2 sin^{3} (u) - 2 cos^{2} (u) sin (u) : 2 sin (u) (sin (u) + cos (u)) (sin (u) - cos (u))

2 sin^{3} (u) - 2 cos^{2} (u) sin (u)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin^{3} (u) = sin (u) sin^{2} (u)

= 2 sin (u) sin^{2} (u) - 2 sin (u) cos^{2} (u)

因式分解出通项

2 sin (u)

= 2 sin (u) (sin^{2} (u) - cos^{2} (u))

分解

sin^{2} (u) - cos^{2} (u) : (sin (u) + cos (u)) (sin (u) - cos (u))

sin^{2} (u) - cos^{2} (u)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

sin^{2} (u) - cos^{2} (u) = (sin (u) + cos (u)) (sin (u) - cos (u))

= (sin (u) + cos (u)) (sin (u) - cos (u))

= 2 sin (u) (sin (u) + cos (u)) (sin (u) - cos (u))

2 sin (u) (sin (u) + cos (u)) (sin (u) - cos (u)) = 0

分别求解每个部分

sin (u) = 0 or sin (u) + cos (u) = 0 or sin (u) - cos (u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 0

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

sin (u) + cos (u) = 0 : u = \frac{3 π}{4} + πn

sin (u) + cos (u) = 0

使用三角恒等式改写

sin (u) + cos (u) = 0

在两边除以

cos (u), cos (u) \neq = 0

\frac{sin ( u ) + cos ( u )}{cos ( u )} = \frac{0}{cos ( u )}

化简

\frac{sin ( u )}{cos ( u )} + 1 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (u) + 1 = 0

tan (u) + 1 = 0

将

1

到右边

tan (u) + 1 = 0

两边减去

1

tan (u) + 1 - 1 = 0 - 1

化简

tan (u) = - 1

tan (u) = - 1

tan (u) = - 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

u = \frac{3 π}{4} + πn

u = \frac{3 π}{4} + πn

sin (u) - cos (u) = 0 : u = \frac{π}{4} + πn

sin (u) - cos (u) = 0

使用三角恒等式改写

sin (u) - cos (u) = 0

在两边除以

cos (u), cos (u) \neq = 0

\frac{sin ( u ) - cos ( u )}{cos ( u )} = \frac{0}{cos ( u )}

化简

\frac{sin ( u )}{cos ( u )} - 1 = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (u) - 1 = 0

tan (u) - 1 = 0

将

1

到右边

tan (u) - 1 = 0

两边加上

1

tan (u) - 1 + 1 = 0 + 1

化简

tan (u) = 1

tan (u) = 1

tan (u) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

u = \frac{π}{4} + πn

u = \frac{π}{4} + πn

合并所有解

u = 2 πn, u = π + 2 πn, u = \frac{3 π}{4} + πn, u = \frac{π}{4} + πn

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = 2 πn, u = π + 2 πn, u = \frac{3 π}{4} + πn, u = \frac{π}{4} + πn

u = \frac{x}{2}

代回

\frac{x}{2} = \frac{π}{2} + 2 πn : x = π + 4 πn

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{2} + 2 \cdot 2 πn : π + 4 πn

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

约分:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

x = π + 4 πn

x = π + 4 πn

x = π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn : x = 3 π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn : 3 π + 4 πn

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 π 2}{2}

约分:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

x = 3 π + 4 πn

\frac{x}{2} = 2 πn : x = 4 πn

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 πn

化简

x = 4 πn

x = 4 πn

\frac{x}{2} = π + 2 πn : x = 2 π + 4 πn

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

化简

x = 2 π + 4 πn

x = 2 π + 4 πn

\frac{x}{2} = \frac{3 π}{4} + πn : x = \frac{3 π}{2} + 2 πn

\frac{x}{2} = \frac{3 π}{4} + πn

在两边乘以

2

\frac{x}{2} = \frac{3 π}{4} + πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{3 π}{4} + 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{3 π}{4} + 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{3 π}{4} + 2 πn : \frac{3 π}{2} + 2 πn

2 \cdot \frac{3 π}{4} + 2 πn

2 \cdot \frac{3 π}{4} = \frac{3 π}{2}

2 \cdot \frac{3 π}{4}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 π 2}{4}

数字相乘:

3 \cdot 2 = 6

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

\frac{x}{2} = \frac{π}{4} + πn : x = \frac{π}{2} + 2 πn

\frac{x}{2} = \frac{π}{4} + πn

在两边乘以

2

\frac{x}{2} = \frac{π}{4} + πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 πn

化简

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 πn

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{4} + 2 πn : \frac{π}{2} + 2 πn

2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = π + 4 πn, x = 3 π + 4 πn, x = 4 πn, x = 2 π + 4 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{2} + 2 πn

作图

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