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受欢迎的三角函数 >

证明 tan(15)=tan(45-30)

解答

证明

tan (1 5^{\circ}) = tan (4 5^{\circ} - 3 0^{\circ})

解答

真

求解步骤

tan (1 5^{\circ}) = tan (4 5^{\circ} - 3 0^{\circ})

调整左侧

tan (1 5^{\circ})

化简

tan (1 5^{\circ}) : 2 - 3

tan (1 5^{\circ})

使用三角恒等式改写:

\frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

tan (1 5^{\circ})

将

tan (1 5^{\circ})

写为

tan (4 5^{\circ} - 3 0^{\circ})

= tan (4 5^{\circ} - 3 0^{\circ})

使用角差恒等式:

tan (s - t) = \frac{tan ( s ) - tan ( t )}{1 + tan ( s ) tan ( t )}

= \frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

= \frac{tan ( 4 5 ^{\circ} ) - tan ( 3 0 ^{\circ} )}{1 + tan ( 4 5 ^{\circ} ) tan ( 3 0 ^{\circ} )}

使用以下普通恒等式:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

使用以下普通恒等式:

tan (3 0^{\circ}) = \frac{3}{3}

tan (3 0^{\circ})

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= \frac{3}{3}

= \frac{1 - \frac{3}{3}}{1 + 1 \cdot \frac{3}{3}}

化简

\frac{1 - \frac{3}{3}}{1 + 1 \cdot \frac{3}{3}} : 2 - 3

\frac{1 - \frac{3}{3}}{1 + 1 \cdot \frac{3}{3}}

乘以:

1 \cdot \frac{3}{3} = \frac{3}{3}

= \frac{1 - \frac{3}{3}}{1 + \frac{3}{3}}

化简

1 + \frac{3}{3} : \frac{3 + 1}{3}

1 + \frac{3}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} + \frac{3}{3}

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{1 \cdot 3 + 3}{3}

数字相乘:

1 \cdot 3 = 3

= \frac{3 + 3}{3}

分解

3 + 3 : 3 (3 + 1)

3 + 3

3 = 33

= 33 + 3

因式分解出通项

3

= 3 (3 + 1)

= \frac{3 ( 3 + 1 )}{3}

消掉

\frac{3 ( 3 + 1 )}{3} : \frac{3 + 1}{3}

\frac{3 ( 3 + 1 )}{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}} ( 1 + 3 )}{3}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{3 + 1}{3 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 + 1}{3 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{3 + 1}{3}

= \frac{3 + 1}{3}

= \frac{1 - \frac{3}{3}}{\frac{3 + 1}{3}}

化简

1 - \frac{3}{3} : \frac{3 - 1}{3}

1 - \frac{3}{3}

将项转换为分式:

1 = \frac{1 \cdot 3}{3}

= \frac{1 \cdot 3}{3} - \frac{3}{3}

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{1 \cdot 3 - 3}{3}

数字相乘:

1 \cdot 3 = 3

= \frac{3 - 3}{3}

分解

3 - 3 : 3 (3 - 1)

3 - 3

3 = 33

= 33 - 3

因式分解出通项

3

= 3 (3 - 1)

= \frac{3 ( 3 - 1 )}{3}

消掉

\frac{3 ( 3 - 1 )}{3} : \frac{3 - 1}{3}

\frac{3 ( 3 - 1 )}{3}

使用根式运算法则:

n a = a^{\frac{1}{n}}

3 = 3^{\frac{1}{2}}

= \frac{3 ^{\frac{1}{2}} ( 3 - 1 )}{3}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{3 ^{\frac{1}{2}}}{3 ^{1}} = \frac{1}{3 ^{1 - \frac{1}{2}}}

= \frac{3 - 1}{3 ^{1 - \frac{1}{2}}}

数字相减:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{3 - 1}{3 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

3^{\frac{1}{2}} = 3

= \frac{3 - 1}{3}

= \frac{3 - 1}{3}

= \frac{\frac{3 - 1}{3}}{\frac{3 + 1}{3}}

分式相除:

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

= \frac{( 3 - 1 ) 3}{3 ( 3 + 1 )}

约分:

3

= \frac{3 - 1}{3 + 1}

\frac{3 - 1}{3 + 1}

有理化:

2 - 3

\frac{3 - 1}{3 + 1}

乘以共轭根式

\frac{3 - 1}{3 - 1}

= \frac{( 3 - 1 ) ( 3 - 1 )}{( 3 + 1 ) ( 3 - 1 )}

(3 - 1) (3 - 1) = 4 - 23

(3 - 1) (3 - 1)

使用指数法则:

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

(3 - 1) (3 - 1) = (3 - 1)^{1 + 1}

= (3 - 1)^{1 + 1}

数字相加:

1 + 1 = 2

= (3 - 1)^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 3, b = 1

= (3)^{2} - 23 \cdot 1 + 1^{2}

化简

(3)^{2} - 23 \cdot 1 + 1^{2} : 4 - 23

(3)^{2} - 23 \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (3)^{2} - 2 \cdot 1 \cdot 3 + 1

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 3

23 \cdot 1 = 23

23 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 23

= 3 - 23 + 1

数字相加:

3 + 1 = 4

= 4 - 23

= 4 - 23

(3 + 1) (3 - 1) = 2

(3 + 1) (3 - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 3, b = 1

= (3)^{2} - 1^{2}

化简

(3)^{2} - 1^{2} : 2

(3)^{2} - 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (3)^{2} - 1

(3)^{2} = 3

(3)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 3

= 3 - 1

数字相减:

3 - 1 = 2

= 2

= 2

= \frac{4 - 2 3}{2}

分解

4 - 23 : 2 (2 - 3)

4 - 23

改写为

= 2 \cdot 2 - 23

因式分解出通项

2

= 2 (2 - 3)

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

数字相除:

\frac{2}{2} = 1

= 2 - 3

= 2 - 3

= 2 - 3

= 2 - 3

调整右侧

tan (4 5^{\circ} - 3 0^{\circ})

使用三角恒等式改写

tan (4 5^{\circ} - 3 0^{\circ})

使用基本三角恒等式:

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

= \frac{sin ( 4 5 ^{\circ} - 3 0 ^{\circ} )}{cos ( 4 5 ^{\circ} - 3 0 ^{\circ} )}

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) - cos ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}{cos ( 4 5 ^{\circ} - 3 0 ^{\circ} )}

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) - cos ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}{cos ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) + sin ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}

\frac{sin ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) - cos ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}{cos ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) + sin ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )} = 2 - 3

\frac{sin ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) - cos ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}{cos ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) + sin ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}

sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ}) = \frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}

sin (4 5^{\circ}) cos (3 0^{\circ}) - cos (4 5^{\circ}) sin (3 0^{\circ})

sin (4 5^{\circ}) cos (3 0^{\circ}) = \frac{2}{2} \cdot \frac{3}{2}

sin (4 5^{\circ}) cos (3 0^{\circ})

化简

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (3 0^{\circ})

化简

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{2}{2} \cdot \frac{3}{2}

cos (4 5^{\circ}) sin (3 0^{\circ}) = \frac{1}{2} \cdot \frac{2}{2}

cos (4 5^{\circ}) sin (3 0^{\circ})

化简

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} sin (3 0^{\circ})

化简

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2} \cdot \frac{2}{2}

= \frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}

= \frac{\frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}}{cos ( 4 5 ^{\circ} ) cos ( 3 0 ^{\circ} ) + sin ( 4 5 ^{\circ} ) sin ( 3 0 ^{\circ} )}

cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ}) = \frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}

cos (4 5^{\circ}) cos (3 0^{\circ}) + sin (4 5^{\circ}) sin (3 0^{\circ})

cos (4 5^{\circ}) cos (3 0^{\circ}) = \frac{2}{2} \cdot \frac{3}{2}

cos (4 5^{\circ}) cos (3 0^{\circ})

化简

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (3 0^{\circ})

化简

cos (3 0^{\circ}) : \frac{3}{2}

cos (3 0^{\circ})

使用以下普通恒等式:

cos (3 0^{\circ}) = \frac{3}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

= \frac{2}{2} \cdot \frac{3}{2}

sin (4 5^{\circ}) sin (3 0^{\circ}) = \frac{1}{2} \cdot \frac{2}{2}

sin (4 5^{\circ}) sin (3 0^{\circ})

化简

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} sin (3 0^{\circ})

化简

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

使用以下普通恒等式:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2} \cdot \frac{2}{2}

= \frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}

= \frac{\frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}}{\frac{2}{2} \cdot \frac{3}{2} + \frac{1}{2} \cdot \frac{2}{2}}

化简

\frac{\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2}}{\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2}}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

分式相乘:

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

= \frac{2 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

\frac{2}{2} \cdot \frac{1}{2}

分式相乘:

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

= \frac{2 \cdot 1}{2 \cdot 2}

乘以:

2 \cdot 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{\frac{2}{2} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{2}{2}}{\frac{6}{4} + \frac{2}{4}}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

分式相乘:

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

= \frac{2 3}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2 3}{4}

化简

23 : 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

\frac{2}{2} \cdot \frac{1}{2}

分式相乘:

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

= \frac{2 \cdot 1}{2 \cdot 2}

乘以:

2 \cdot 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{\frac{6}{4} - \frac{2}{4}}{\frac{6}{4} + \frac{2}{4}}

合并分式

\frac{6}{4} + \frac{2}{4} : \frac{6 + 2}{4}

使用法则

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

= \frac{6 + 2}{4}

= \frac{\frac{6}{4} - \frac{2}{4}}{\frac{6 + 2}{4}}

合并分式

\frac{6}{4} - \frac{2}{4} : \frac{6 - 2}{4}

使用法则

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

= \frac{6 - 2}{4}

= \frac{\frac{6 - 2}{4}}{\frac{6 + 2}{4}}

分式相除:

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

= \frac{( 6 - 2 ) \cdot 4}{4 ( 6 + 2 )}

约分:

4

= \frac{6 - 2}{6 + 2}

\frac{6 - 2}{6 + 2}

有理化:

2 - 3

\frac{6 - 2}{6 + 2}

乘以共轭根式

\frac{6 - 2}{6 - 2}

= \frac{( 6 - 2 ) ( 6 - 2 )}{( 6 + 2 ) ( 6 - 2 )}

(6 - 2) (6 - 2) = 8 - 43

(6 - 2) (6 - 2)

使用指数法则:

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

(6 - 2) (6 - 2) = (6 - 2)^{1 + 1}

= (6 - 2)^{1 + 1}

数字相加:

1 + 1 = 2

= (6 - 2)^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 6, b = 2

= (6)^{2} - 262 + (2)^{2}

化简

(6)^{2} - 262 + (2)^{2} : 8 - 43

(6)^{2} - 262 + (2)^{2}

(6)^{2} = 6

(6)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 6

262 = 43

262

分解整数

6 = 2 \cdot 3

= 2 2 \cdot 3 2

使用根式运算法则:

n ab = n a n b

2 \cdot 3 = 23

= 2232

使用根式运算法则:

a a = a

22 = 2

= 2 \cdot 23

数字相乘:

2 \cdot 2 = 4

= 43

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 6 - 43 + 2

数字相加:

6 + 2 = 8

= 8 - 43

= 8 - 43

(6 + 2) (6 - 2) = 4

(6 + 2) (6 - 2)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 6, b = 2

= (6)^{2} - (2)^{2}

化简

(6)^{2} - (2)^{2} : 4

(6)^{2} - (2)^{2}

(6)^{2} = 6

(6)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (6^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 6^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 6

(2)^{2} = 2

(2)^{2}

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 6 - 2

数字相减:

6 - 2 = 4

= 4

= 4

= \frac{8 - 4 3}{4}

分解

8 - 43 : 4 (2 - 3)

8 - 43

改写为

= 4 \cdot 2 - 43

因式分解出通项

4

= 4 (2 - 3)

= \frac{4 ( 2 - 3 )}{4}

数字相除:

\frac{4}{4} = 1

= 2 - 3

= 2 - 3

= 2 - 3

= 2 - 3

= 2 - 3

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 sec(x)sin(2x)=2sin(x)

p ro v e sec (x) sin (2 x) = 2 sin (x)

证明 sec(x)cot(x)-cot(x)cos(x)=sin(x)

p ro v e sec (x) cot (x) - cot (x) cos (x) = sin (x)

证明 cos(x)*cot(x)=csc(x)-sin(x)

p ro v e cos (x) \cdot cot (x) = csc (x) - sin (x)

证明 cos(x+270)=sin(x)

p ro v e cos (x + 27 0^{\circ}) = sin (x)

证明 sin(θ)cos(θ)sec(θ)csc(θ)=1

p ro v e sin (θ) cos (θ) sec (θ) csc (θ) = 1