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受欢迎的三角函数 >

arctan(0.1x)+arctan(0.01x)=39

解答

arctan (0.1 x) + arctan (0.01 x) = 3 9^{\circ}

解答

x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

求解步骤

arctan (0.1 x) + arctan (0.01 x) = 3 9^{\circ}

使用三角恒等式改写

arctan (0.1 x) + arctan (0.01 x)

使用和差化积恒等式:

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

= arctan (\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x})

arctan (\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x}) = 3 9^{\circ}

使用反三角函数性质

arctan (\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x}) = 3 9^{\circ}

arctan (x) = a \Rightarrow x = tan (a)

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x} = tan (3 9^{\circ})

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x} = tan (3 9^{\circ})

解

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x} = tan (3 9^{\circ}) : x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x} = tan (3 9^{\circ})

化简

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x} : \frac{0.11 x}{1 - 0.001 x ^{2}}

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x}

同类项相加:

0.1 x + 0.01 x = 0.11 x

= \frac{0.11 x}{1 - 0.1 \cdot 0.01 xx}

1 - 0.1 x \cdot 0.01 x = 1 - 0.001 x^{2}

1 - 0.1 x \cdot 0.01 x

0.1 x \cdot 0.01 x = 0.001 x^{2}

0.1 x \cdot 0.01 x

数字相乘:

0.1 \cdot 0.01 = 0.001

= 0.001 xx

使用指数法则:

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

xx = x^{1 + 1}

= 0.001 x^{1 + 1}

数字相加:

1 + 1 = 2

= 0.001 x^{2}

= 1 - 0.001 x^{2}

= \frac{0.11 x}{1 - 0.001 x ^{2}}

\frac{0.11 x}{1 - 0.001 x ^{2}} = tan (3 9^{\circ})

在两边乘以

1 - 0.001 x^{2}

\frac{0.11 x}{1 - 0.001 x ^{2}} = tan (3 9^{\circ})

在两边乘以

1 - 0.001 x^{2}

\frac{0.11 x}{1 - 0.001 x ^{2}} (1 - 0.001 x^{2}) = tan (3 9^{\circ}) (1 - 0.001 x^{2})

化简

0.11 x = tan (3 9^{\circ}) (1 - 0.001 x^{2})

0.11 x = tan (3 9^{\circ}) (1 - 0.001 x^{2})

解

0.11 x = tan (3 9^{\circ}) (1 - 0.001 x^{2}) : x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

0.11 x = tan (3 9^{\circ}) (1 - 0.001 x^{2})

展开

tan (3 9^{\circ}) (1 - 0.001 x^{2}) : tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2}

tan (3 9^{\circ}) (1 - 0.001 x^{2})

使用分配律:

a (b - c) = ab - a c

a = tan (3 9^{\circ}), b = 1, c = 0.001 x^{2}

= tan (3 9^{\circ}) \cdot 1 - tan (3 9^{\circ}) \cdot 0.001 x^{2}

= 1 \cdot tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2}

乘以:

1 \cdot tan (3 9^{\circ}) = tan (3 9^{\circ})

= tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2}

0.11 x = tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2}

交换两边

tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2} = 0.11 x

将

0.11 x

para o lado esquerdo

tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2} = 0.11 x

两边减去

0.11 x

tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2} - 0.11 x = 0.11 x - 0.11 x

化简

tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2} - 0.11 x = 0

tan (3 9^{\circ}) - 0.001 tan (3 9^{\circ}) x^{2} - 0.11 x = 0

改写成标准形式

a x^{2} + b x + c = 0

- 0.00080 \dots x^{2} - 0.11 x + 0.80978 \dots = 0

使用求根公式求解

- 0.00080 \dots x^{2} - 0.11 x + 0.80978 \dots = 0

二次方程求根公式:

若

a = - 0.00080 \dots, b = - 0.11, c = 0.80978 \dots

x_{1, 2} = \frac{- ( - 0.11 ) \pm ( - 0.11 ) ^{2} - 4 ( - 0.00080 \dots ) \cdot 0.80978 \dots}{2 ( - 0.00080 \dots )}

x_{1, 2} = \frac{- ( - 0.11 ) \pm ( - 0.11 ) ^{2} - 4 ( - 0.00080 \dots ) \cdot 0.80978 \dots}{2 ( - 0.00080 \dots )}

(- 0.11)^{2} - 4 (- 0.00080 \dots) \cdot 0.80978 \dots = 0.01472 \dots

(- 0.11)^{2} - 4 (- 0.00080 \dots) \cdot 0.80978 \dots

使用法则

- (- a) = a

= (- 0.11)^{2} + 4 \cdot 0.00080 \dots \cdot 0.80978 \dots

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 0.11)^{2} = 0.1 1^{2}

= 0.1 1^{2} + 4 \cdot 0.00080 \dots \cdot 0.80978 \dots

数字相乘:

4 \cdot 0.00080 \dots \cdot 0.80978 \dots = 0.00262 \dots

= 0.1 1^{2} + 0.00262 \dots

0.1 1^{2} = 0.0121

= 0.0121 + 0.00262 \dots

数字相加:

0.0121 + 0.00262 \dots = 0.01472 \dots

= 0.01472 \dots

x_{1, 2} = \frac{- ( - 0.11 ) \pm 0.01472 \dots}{2 ( - 0.00080 \dots )}

将解分隔开

x_{1} = \frac{- ( - 0.11 ) + 0.01472 \dots}{2 ( - 0.00080 \dots )}, x_{2} = \frac{- ( - 0.11 ) - 0.01472 \dots}{2 ( - 0.00080 \dots )}

x = \frac{- ( - 0.11 ) + 0.01472 \dots}{2 ( - 0.00080 \dots )} : - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

\frac{- ( - 0.11 ) + 0.01472 \dots}{2 ( - 0.00080 \dots )}

去除括号:

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

= \frac{0.11 + 0.01472 \dots}{- 2 \cdot 0.00080 \dots}

数字相乘:

2 \cdot 0.00080 \dots = 0.00161 \dots

= \frac{0.11 + 0.01472 \dots}{- 0.00161 \dots}

使用分式法则:

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

= - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

x = \frac{- ( - 0.11 ) - 0.01472 \dots}{2 ( - 0.00080 \dots )} : \frac{0.01472 \dots - 0.11}{0.00161 \dots}

\frac{- ( - 0.11 ) - 0.01472 \dots}{2 ( - 0.00080 \dots )}

去除括号:

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

= \frac{0.11 - 0.01472 \dots}{- 2 \cdot 0.00080 \dots}

数字相乘:

2 \cdot 0.00080 \dots = 0.00161 \dots

= \frac{0.11 - 0.01472 \dots}{- 0.00161 \dots}

使用分式法则:

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

0.11 - 0.01472 \dots = - (0.01472 \dots - 0.11)

= \frac{0.01472 \dots - 0.11}{0.00161 \dots}

二次方程组的解是:

x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

验证解

找到无定义的点（奇点）:

x = 1010, x = - 1010

取

\frac{0.1 x + 0.01 x}{1 - 0.1 x \cdot 0.01 x}

的分母，令其等于零

解

1 - 0.1 x \cdot 0.01 x = 0 : x = 1010, x = - 1010

1 - 0.1 x \cdot 0.01 x = 0

将

1

到右边

1 - 0.1 x \cdot 0.01 x = 0

两边减去

1

1 - 0.1 x \cdot 0.01 x - 1 = 0 - 1

化简

- 0.1 x \cdot 0.01 x = - 1

- 0.1 x \cdot 0.01 x = - 1

化简

- 0.001 x^{2} = - 1

两边除以

- 0.001

\frac{- 0.001 x ^{2}}{- 0.001} = \frac{- 1}{- 0.001}

x^{2} = \frac{1}{0.001}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

x = \frac{1}{0.001}, x = - \frac{1}{0.001}

\frac{1}{0.001} = 1010

\frac{1}{0.001}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{0.001}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{0.001}

0.001 = \frac{1}{10 10}

0.001

0.001 = \frac{1}{1000}

0.001

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

3

位，因此乘以和除以

1000

= \frac{1000 \cdot 0.001}{1000}

数字相乘:

1000 \cdot 0.001 = 1

= \frac{1}{1000}

= \frac{1}{1000}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{1000}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{1000}

1000 = 1010

1000

1000

质因数分解:

2^{3} \cdot 5^{3}

1000

1000

除以

21000 = 500 \cdot 2

= 2 \cdot 500

500

除以

2500 = 250 \cdot 2

= 2 \cdot 2 \cdot 250

250

除以

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 125

125

除以

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

= 2^{3} \cdot 5^{3}

= 2^{3} \cdot 5^{3}

使用指数法则:

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

2^{3} \cdot 5^{3} = 2^{2} \cdot 2 \cdot 5^{2} \cdot 5

= 2^{2} \cdot 2 \cdot 5^{2} \cdot 5

2^{2} \cdot 2 \cdot 5^{2} \cdot 5 = 2^{2} 5^{2} 2 \cdot 5

2^{2} \cdot 2 \cdot 5^{2} \cdot 5

使用根式运算法则:

ab = a b, a \geq 0, b \geq 0

2^{2} \cdot 2 \cdot 5^{2} \cdot 5 = 2^{2} 2 \cdot 5^{2} \cdot 5

= 2^{2} 2 \cdot 5^{2} \cdot 5

使用根式运算法则:

ab = a b, a \geq 0, b \geq 0

2 \cdot 5^{2} \cdot 5 = 5^{2} 2 \cdot 5

= 2^{2} 5^{2} 2 \cdot 5

= 2^{2} 5^{2} 2 \cdot 5

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2 5^{2} 2 \cdot 5

使用根式运算法则:

a^{2} = a, a \geq 0

5^{2} = 5

= 2 \cdot 5 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= 1010

= \frac{1}{10 10}

= \frac{1}{\frac{1}{10 10}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{10 10}{1}

使用分式法则:

\frac{a}{1} = a

= 1010

- \frac{1}{0.001} = - 1010

- \frac{1}{0.001}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{0.001}

使用根式运算法则:

1 = 1

1 = 1

= - \frac{1}{0.001}

0.001 = \frac{1}{10 10}

0.001

0.001 = \frac{1}{1000}

0.001

小数点后有几位数字就乘以和除以几次 10。

小数点右侧有

3

位，因此乘以和除以

1000

= \frac{1000 \cdot 0.001}{1000}

数字相乘:

1000 \cdot 0.001 = 1

= \frac{1}{1000}

= \frac{1}{1000}

使用根式运算法则:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{1000}

使用根式运算法则:

1 = 1

1 = 1

= \frac{1}{1000}

1000 = 1010

1000

1000

质因数分解:

2^{3} \cdot 5^{3}

1000

1000

除以

21000 = 500 \cdot 2

= 2 \cdot 500

500

除以

2500 = 250 \cdot 2

= 2 \cdot 2 \cdot 250

250

除以

2250 = 125 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 125

125

除以

5125 = 25 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5

= 2^{3} \cdot 5^{3}

= 2^{3} \cdot 5^{3}

使用指数法则:

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

2^{3} \cdot 5^{3} = 2^{2} \cdot 2 \cdot 5^{2} \cdot 5

= 2^{2} \cdot 2 \cdot 5^{2} \cdot 5

2^{2} \cdot 2 \cdot 5^{2} \cdot 5 = 2^{2} 5^{2} 2 \cdot 5

2^{2} \cdot 2 \cdot 5^{2} \cdot 5

使用根式运算法则:

ab = a b, a \geq 0, b \geq 0

2^{2} \cdot 2 \cdot 5^{2} \cdot 5 = 2^{2} 2 \cdot 5^{2} \cdot 5

= 2^{2} 2 \cdot 5^{2} \cdot 5

使用根式运算法则:

ab = a b, a \geq 0, b \geq 0

2 \cdot 5^{2} \cdot 5 = 5^{2} 2 \cdot 5

= 2^{2} 5^{2} 2 \cdot 5

= 2^{2} 5^{2} 2 \cdot 5

使用根式运算法则:

a^{2} = a, a \geq 0

2^{2} = 2

= 2 5^{2} 2 \cdot 5

使用根式运算法则:

a^{2} = a, a \geq 0

5^{2} = 5

= 2 \cdot 5 2 \cdot 5

数字相乘:

2 \cdot 5 = 10

= 1010

= \frac{1}{10 10}

= - \frac{1}{\frac{1}{10 10}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

\frac{1}{\frac{1}{10 10}} = \frac{10 10}{1}

= - \frac{10 10}{1}

使用分式法则:

\frac{a}{1} = a

= - 1010

x = 1010, x = - 1010

以下点无定义

x = 1010, x = - 1010

将不在定义域的点与解相综合:

x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}, x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

将解代入原方程进行验证

将它们代入

arctan (0.1 x) + arctan (0.01 x) = 3 9^{\circ}

检验解是否符合

去除与方程不符的解。

检验

- \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

的解:

假

- \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

代入

n = 1

- \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

对于

arctan (0.1 x) + arctan (0.01 x) = 3 9^{\circ}

代入

x = - \frac{0.11 + 0.01472 \dots}{0.00161 \dots}

arctan (0.1 (- \frac{0.11 + 0.01472 \dots}{0.00161 \dots})) + arctan (0.01 (- \frac{0.11 + 0.01472 \dots}{0.00161 \dots})) = 3 9^{\circ}

整理后得

- 2.46091 \dots = 0.68067 \dots

\Rightarrow 假

检验

\frac{0.01472 \dots - 0.11}{0.00161 \dots}

的解:

真

\frac{0.01472 \dots - 0.11}{0.00161 \dots}

代入

n = 1

\frac{0.01472 \dots - 0.11}{0.00161 \dots}

对于

arctan (0.1 x) + arctan (0.01 x) = 3 9^{\circ}

代入

x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

arctan (0.1 \cdot \frac{0.01472 \dots - 0.11}{0.00161 \dots}) + arctan (0.01 \cdot \frac{0.01472 \dots - 0.11}{0.00161 \dots}) = 3 9^{\circ}

整理后得

0.68067 \dots = 0.68067 \dots

\Rightarrow 真

x = \frac{0.01472 \dots - 0.11}{0.00161 \dots}

作图

流行的例子

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sin(c)=(5sin(140))/7

sin (c) = \frac{5 sin ( 14 0 ^{\circ} )}{7}

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2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, θ

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cos^{2} (3 x) - sin^{2} (3 x) = - 1