zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tan(112.5)

解答

tan (112. 5^{\circ})

解答

- 3 + 22

+1

十进制

- 2.41421 \dots

求解步骤

tan (112. 5^{\circ})

使用三角恒等式改写:

- \frac{1 - cos ( 22 5 ^{\circ} )}{1 + cos ( 22 5 ^{\circ} )}

tan (112. 5^{\circ})

将

tan (112. 5^{\circ})

写为

tan (\frac{22 5 ^{\circ}}{2})

= tan (\frac{22 5 ^{\circ}}{2})

使用半角公式:

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

使用三角恒等式改写:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

利用以下特性

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

两边进行平方

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

使用三角恒等式改写:

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

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

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

使用三角恒等式改写:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

化简

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

用

\frac{θ}{2}

替代

θ

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( 2 \cdot \frac{θ}{2} )}{1 + cos ( 2 \cdot \frac{θ}{2} )}

化简

tan^{2} (\frac{θ}{2}) = \frac{1 - cos ( θ )}{1 + cos ( θ )}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

:

range [0, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] quadrant I II tan positive negative

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

= - \frac{1 - cos ( 22 5 ^{\circ} )}{1 + cos ( 22 5 ^{\circ} )}

= - \frac{1 - cos ( 22 5 ^{\circ} )}{1 + cos ( 22 5 ^{\circ} )}

使用三角恒等式改写:

cos (22 5^{\circ}) = - \frac{2}{2}

cos (22 5^{\circ})

使用三角恒等式改写:

cos (18 0^{\circ}) cos (4 5^{\circ}) - sin (18 0^{\circ}) sin (4 5^{\circ})

cos (22 5^{\circ})

将

cos (22 5^{\circ})

写为

cos (18 0^{\circ} + 4 5^{\circ})

= cos (18 0^{\circ} + 4 5^{\circ})

使用角和恒等式:

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

= cos (18 0^{\circ}) cos (4 5^{\circ}) - sin (18 0^{\circ}) sin (4 5^{\circ})

= cos (18 0^{\circ}) cos (4 5^{\circ}) - sin (18 0^{\circ}) sin (4 5^{\circ})

使用以下普通恒等式:

cos (18 0^{\circ}) = (- 1)

cos (18 0^{\circ})

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}

= (- 1)

使用以下普通恒等式:

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

cos (4 5^{\circ})

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}

使用以下普通恒等式:

sin (18 0^{\circ}) = 0

sin (18 0^{\circ})

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}

= 0

使用以下普通恒等式:

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

sin (4 5^{\circ})

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}

= (- 1) \frac{2}{2} - 0 \cdot \frac{2}{2}

化简

= - \frac{2}{2}

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

化简

- \frac{1 - ( - \frac{2}{2} )}{1 - \frac{2}{2}} : - 3 + 22

- \frac{1 - ( - \frac{2}{2} )}{1 - \frac{2}{2}}

使用法则

- (- a) = a

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

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

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

化简

1 - \frac{2}{2} : \frac{2 - 2}{2}

1 - \frac{2}{2}

将项转换为分式:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2 - 2}{2}

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

化简

1 + \frac{2}{2} : \frac{2 + 2}{2}

1 + \frac{2}{2}

将项转换为分式:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2 + 2}{2}

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

分式相除:

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

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

约分:

2

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

分解

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

因式分解出通项

2

= 2 (2 + 1)

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

分解

2 - 2 : 2 (2 - 1)

2 - 2

2 = 22

= 22 - 2

因式分解出通项

2

= 2 (2 - 1)

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

约分:

2

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

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

\frac{2 + 1}{2 - 1} = 3 + 22

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

乘以共轭根式

\frac{2 + 1}{2 + 1}

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

(2 + 1) (2 + 1) = 3 + 22

(2 + 1) (2 + 1)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= (2 + 1)^{2}

使用完全平方公式:

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

a = 2, b = 1

= (2)^{2} + 22 \cdot 1 + 1^{2}

化简

(2)^{2} + 22 \cdot 1 + 1^{2} : 3 + 22

(2)^{2} + 22 \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

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

(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

22 \cdot 1 = 22

22 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 22

= 2 + 22 + 1

数字相加:

2 + 1 = 3

= 3 + 22

= 3 + 22

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

(2 - 1) (2 + 1)

使用平方差公式:

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

a = 2, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(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

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= 1

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

使用法则

\frac{a}{1} = a

= 3 + 22

= - 3 + 22

= - 3 + 22

流行的例子

tan(arcsin((sqrt(3))/2))

tan (arcsin (\frac{3}{2}))

tan (5 8^{\circ})

tan (6 5^{\circ})

tan (- 15 0^{\circ})

tan (4 3^{\circ})