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受欢迎的三角函数 >

tan(37.5)

解答

tan (37. 5^{\circ})

解答

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

+1

十进制

0.76732 \dots

求解步骤

tan (37. 5^{\circ})

使用三角恒等式改写:

\frac{1 - cos ( 7 5 ^{\circ} )}{1 + cos ( 7 5 ^{\circ} )}

tan (37. 5^{\circ})

将

tan (37. 5^{\circ})

写为

tan (\frac{7 5 ^{\circ}}{2})

= tan (\frac{7 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 ( 7 5 ^{\circ} )}{1 + cos ( 7 5 ^{\circ} )}

= \frac{1 - cos ( 7 5 ^{\circ} )}{1 + cos ( 7 5 ^{\circ} )}

使用三角恒等式改写:

cos (7 5^{\circ}) = \frac{6 - 2}{4}

cos (7 5^{\circ})

使用三角恒等式改写:

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

cos (7 5^{\circ})

将

cos (7 5^{\circ})

写为

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

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

使用角和恒等式:

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

= 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})

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:

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

cos (3 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}

= \frac{3}{2}

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}

使用以下普通恒等式:

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

sin (3 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}

= \frac{1}{2}

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

化简

\frac{2}{2} \cdot \frac{3}{2} - \frac{2}{2} \cdot \frac{1}{2} : \frac{6 - 2}{4}

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

使用法则

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

= \frac{6 - 2}{4}

= \frac{6 - 2}{4}

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

化简

\frac{1 - \frac{6 - 2}{4}}{1 + \frac{6 - 2}{4}} : \frac{4 - 6 + 2}{4 + 6 - 2}

\frac{1 - \frac{6 - 2}{4}}{1 + \frac{6 - 2}{4}}

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

\frac{1 - \frac{6 - 2}{4}}{1 + \frac{6 - 2}{4}}

化简

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

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

将项转换为分式:

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

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

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 4 + 6 - 2}{4}

数字相乘:

1 \cdot 4 = 4

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

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

化简

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

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

将项转换为分式:

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

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

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 4 - ( 6 - 2 )}{4}

数字相乘:

1 \cdot 4 = 4

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

- (6 - 2) : - 6 + 2

- (6 - 2)

打开括号

= - (6) - (- 2)

使用加减运算法则

- (- a) = a, - (a) = - a

= - 6 + 2

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

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

分式相除:

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

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

约分:

4

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

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

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

流行的例子

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

60 \cdot cos (3 7^{\circ})

45 sin (7 0^{\circ})

sec^2(45)+csc(30)

sec^{2} (4 5^{\circ}) + csc (3 0^{\circ})

240 \cdot cos (3 0^{\circ})