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受欢迎的三角函数 >

tan(θ/2)-sin(θ)=0

解答

tan (\frac{θ}{2}) - sin (θ) = 0

解答

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

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 72 0^{\circ} n, θ = 36 0^{\circ} + 72 0^{\circ} n

求解步骤

tan (\frac{θ}{2}) - sin (θ) = 0

用 sin, cos 表示

- sin (θ) + tan (\frac{θ}{2})

使用基本三角恒等式:

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

= - sin (θ) + \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

化简

- sin (θ) + \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )} : \frac{- sin ( θ ) cos ( \frac{θ}{2} ) + sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

- sin (θ) + \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

将项转换为分式:

sin (θ) = \frac{sin ( θ ) cos ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

= - \frac{sin ( θ ) cos ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )} + \frac{sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

因为分母相等，所以合并分式:

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

= \frac{- sin ( θ ) cos ( \frac{θ}{2} ) + sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

= \frac{- sin ( θ ) cos ( \frac{θ}{2} ) + sin ( \frac{θ}{2} )}{cos ( \frac{θ}{2} )}

\frac{sin ( \frac{θ}{2} ) - cos ( \frac{θ}{2} ) sin ( θ )}{cos ( \frac{θ}{2} )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (\frac{θ}{2}) - cos (\frac{θ}{2}) sin (θ) = 0

使用三角恒等式改写

sin (\frac{θ}{2}) - cos (\frac{θ}{2}) sin (θ)

使用积化和差恒等式:

sin (s) cos (t) = \frac{1}{2} (sin (s + t) + sin (s - t))

= sin (\frac{θ}{2}) - \frac{1}{2} (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

化简

sin (\frac{θ}{2}) - \frac{1}{2} (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})) : \frac{- sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

sin (\frac{θ}{2}) - \frac{1}{2} (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

\frac{1}{2} (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})) = \frac{sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

\frac{1}{2} (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

分式相乘:

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

= \frac{1 \cdot ( sin ( θ + \frac{θ}{2} ) + sin ( θ - \frac{θ}{2} ) )}{2}

1 \cdot (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})) = sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})

1 \cdot (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

乘以:

1 \cdot (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})) = (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

= (sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2}))

去除括号:

(a) = a

= sin (θ + \frac{θ}{2}) + sin (θ - \frac{θ}{2})

= \frac{sin ( θ + \frac{θ}{2} ) + sin ( θ - \frac{θ}{2} )}{2}

化简

θ + \frac{θ}{2} : \frac{3 θ}{2}

θ + \frac{θ}{2}

将项转换为分式:

θ = \frac{θ 2}{2}

= \frac{θ}{2} + \frac{θ \cdot 2}{2}

因为分母相等，所以合并分式:

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

= \frac{θ + θ \cdot 2}{2}

同类项相加:

θ + 2 θ = 3 θ

= \frac{3 θ}{2}

= \frac{sin ( \frac{3 θ}{2} ) + sin ( θ - \frac{θ}{2} )}{2}

化简

θ - \frac{θ}{2} : \frac{θ}{2}

θ - \frac{θ}{2}

将项转换为分式:

θ = \frac{θ 2}{2}

= - \frac{θ}{2} + \frac{θ \cdot 2}{2}

因为分母相等，所以合并分式:

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

= \frac{- θ + θ \cdot 2}{2}

同类项相加:

- θ + 2 θ = θ

= \frac{θ}{2}

= \frac{sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

= sin (\frac{θ}{2}) - \frac{sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

将项转换为分式:

sin (\frac{θ}{2}) = \frac{sin ( \frac{θ}{2} ) 2}{2}

= - \frac{sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2} + \frac{sin ( \frac{θ}{2} ) \cdot 2}{2}

因为分母相等，所以合并分式:

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

= \frac{- ( sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} ) ) + sin ( \frac{θ}{2} ) \cdot 2}{2}

乘开

- (sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})) + sin (\frac{θ}{2}) \cdot 2 : - sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})

- (sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})) + sin (\frac{θ}{2}) \cdot 2

= - (sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})) + 2 sin (\frac{θ}{2})

- (sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})) : - sin (\frac{3 θ}{2}) - sin (\frac{θ}{2})

- (sin (\frac{3 θ}{2}) + sin (\frac{θ}{2}))

打开括号

= - (sin (\frac{3 θ}{2})) - (sin (\frac{θ}{2}))

使用加减运算法则

+ (- a) = - a

= - sin (\frac{3 θ}{2}) - sin (\frac{θ}{2})

= - sin (\frac{3 θ}{2}) - sin (\frac{θ}{2}) + sin (\frac{θ}{2}) \cdot 2

同类项相加:

- sin (\frac{θ}{2}) + 2 sin (\frac{θ}{2}) = sin (\frac{θ}{2})

= - sin (\frac{3 θ}{2}) + sin (\frac{θ}{2})

= \frac{- sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

= \frac{- sin ( \frac{3 θ}{2} ) + sin ( \frac{θ}{2} )}{2}

使用和差化积恒等式:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

= \frac{2 sin ( \frac{\frac{θ}{2} - \frac{3 θ}{2}}{2} ) cos ( \frac{\frac{θ}{2} + \frac{3 θ}{2}}{2} )}{2}

化简

\frac{2 sin ( \frac{\frac{θ}{2} - \frac{3 θ}{2}}{2} ) cos ( \frac{\frac{θ}{2} + \frac{3 θ}{2}}{2} )}{2} : - cos (θ) sin (\frac{θ}{2})

\frac{2 sin ( \frac{\frac{θ}{2} - \frac{3 θ}{2}}{2} ) cos ( \frac{\frac{θ}{2} + \frac{3 θ}{2}}{2} )}{2}

合并分式

\frac{θ}{2} + \frac{3 θ}{2} : 2 θ

使用法则

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

= \frac{θ + 3 θ}{2}

同类项相加:

θ + 3 θ = 4 θ

= \frac{4 θ}{2}

数字相除:

\frac{4}{2} = 2

= 2 θ

= \frac{2 sin ( \frac{\frac{θ}{2} - \frac{3 θ}{2}}{2} ) cos ( \frac{2 θ}{2} )}{2}

合并分式

\frac{θ}{2} - \frac{3 θ}{2} : - θ

使用法则

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

= \frac{θ - 3 θ}{2}

同类项相加:

θ - 3 θ = - 2 θ

= \frac{- 2 θ}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= - θ

= \frac{2 sin ( \frac{- θ}{2} ) cos ( \frac{2 θ}{2} )}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= \frac{2 sin ( - \frac{θ}{2} ) cos ( \frac{2 θ}{2} )}{2}

数字相除:

\frac{2}{2} = 1

= sin (- \frac{θ}{2}) cos (\frac{2 θ}{2})

使用负角恒等式:

sin (- x) = - sin (x)

= cos (\frac{2 θ}{2}) (- sin (\frac{θ}{2}))

去除括号:

(- a) = - a

= - cos (\frac{2 θ}{2}) sin (\frac{θ}{2})

数字相除:

\frac{2}{2} = 1

= - cos (θ) sin (\frac{θ}{2})

= - cos (θ) sin (\frac{θ}{2})

- cos (θ) sin (\frac{θ}{2}) = 0

分别求解每个部分

cos (θ) = 0 or sin (\frac{θ}{2}) = 0

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

cos (θ) = 0

cos (θ) = 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}

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

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

sin (\frac{θ}{2}) = 0 : θ = 4 πn, θ = 2 π + 4 πn

sin (\frac{θ}{2}) = 0

sin (\frac{θ}{2}) = 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}

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

\frac{θ}{2} = 0 + 2 πn, \frac{θ}{2} = π + 2 πn

解

\frac{θ}{2} = 0 + 2 πn : θ = 4 πn

\frac{θ}{2} = 0 + 2 πn

0 + 2 πn = 2 πn

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{θ}{2} = 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot 2 πn

化简

θ = 4 πn

θ = 4 πn

解

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

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{θ}{2} = π + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 π + 2 \cdot 2 πn

化简

θ = 2 π + 4 πn

θ = 2 π + 4 πn

θ = 4 πn, θ = 2 π + 4 πn

合并所有解

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

作图

流行的例子

-5cos(x)=-2sin^2(x)+4

- 5 cos (x) = - 2 sin^{2} (x) + 4

cos(θ)cos(2θ)+sin(θ)sin(2θ)=(sqrt(2))/2

cos (θ) cos (2 θ) + sin (θ) sin (2 θ) = \frac{2}{2}

sec^2(x)-2tan(x)=4

sec^{2} (x) - 2 tan (x) = 4

cos(4x)=cos(2x)

cos (4 x) = cos (2 x)

sin(4x)+sin(2x)=0

sin (4 x) + sin (2 x) = 0