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受欢迎的三角函数 >

5tan(y)-1= 1/(5tan(y))

解答

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

解答

y = 0.31297 \dots + πn, y = - 0.12298 \dots + πn

+1

度数

y = 17.93193 \dots^{\circ} + 18 0^{\circ} n, y = - 7.04640 \dots^{\circ} + 18 0^{\circ} n

求解步骤

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

用替代法求解

5 tan (y) - 1 = \frac{1}{5 tan ( y )}

令

: tan (y) = u

5 u - 1 = \frac{1}{5 u}

5 u - 1 = \frac{1}{5 u} : u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

5 u - 1 = \frac{1}{5 u}

在两边乘以

u

5 u - 1 = \frac{1}{5 u}

在两边乘以

u

5 uu - 1 \cdot u = \frac{1}{5 u} u

化简

5 uu : 5 u^{2}

5 uu - 1 \cdot u = \frac{1}{5 u} u

使用指数法则:

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

uu = u^{1 + 1}

= 5 u^{1 + 1}

数字相加:

1 + 1 = 2

= 5 u^{2}

5 u^{2} - u = \frac{1}{5}

5 u^{2} - u = \frac{1}{5}

解

5 u^{2} - u = \frac{1}{5} : u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

5 u^{2} - u = \frac{1}{5}

在两边乘以

5

5 u^{2} - u = \frac{1}{5}

在两边乘以

5

5 u^{2} \cdot 5 - u \cdot 5 = \frac{1}{5} \cdot 5

化简

25 u^{2} - 5 u = 1

25 u^{2} - 5 u = 1

将

1

para o lado esquerdo

25 u^{2} - 5 u = 1

两边减去

1

25 u^{2} - 5 u - 1 = 1 - 1

化简

25 u^{2} - 5 u - 1 = 0

25 u^{2} - 5 u - 1 = 0

使用求根公式求解

25 u^{2} - 5 u - 1 = 0

二次方程求根公式:

若

a = 25, b = - 5, c = - 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 25 ( - 1 )}{2 \cdot 25}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 25 ( - 1 )}{2 \cdot 25}

(- 5)^{2} - 4 \cdot 25 (- 1) = 55

(- 5)^{2} - 4 \cdot 25 (- 1)

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 25 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 25 \cdot 1

数字相乘:

4 \cdot 25 \cdot 1 = 100

= 5^{2} + 100

5^{2} = 25

= 25 + 100

数字相加:

25 + 100 = 125

= 125

125

质因数分解:

5^{3}

125

125

除以

5125 = 25 \cdot 5

= 5 \cdot 25

25

除以

525 = 5 \cdot 5

= 5 \cdot 5 \cdot 5

5

是质数，因此无法进一步因数分解

= 5 \cdot 5 \cdot 5

= 5^{3}

= 5^{3}

使用指数法则:

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

= 5^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 55

u_{1, 2} = \frac{- ( - 5 ) \pm 5 5}{2 \cdot 25}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 5 5}{2 \cdot 25}, u_{2} = \frac{- ( - 5 ) - 5 5}{2 \cdot 25}

u = \frac{- ( - 5 ) + 5 5}{2 \cdot 25} : \frac{1 + 5}{10}

\frac{- ( - 5 ) + 5 5}{2 \cdot 25}

使用法则

- (- a) = a

= \frac{5 + 5 5}{2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{5 + 5 5}{50}

分解

5 + 55 : 5 (1 + 5)

5 + 55

改写为

= 5 \cdot 1 + 55

因式分解出通项

5

= 5 (1 + 5)

= \frac{5 ( 1 + 5 )}{50}

约分:

5

= \frac{1 + 5}{10}

u = \frac{- ( - 5 ) - 5 5}{2 \cdot 25} : \frac{1 - 5}{10}

\frac{- ( - 5 ) - 5 5}{2 \cdot 25}

使用法则

- (- a) = a

= \frac{5 - 5 5}{2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{5 - 5 5}{50}

分解

5 - 55 : 5 (1 - 5)

5 - 55

改写为

= 5 \cdot 1 - 55

因式分解出通项

5

= 5 (1 - 5)

= \frac{5 ( 1 - 5 )}{50}

约分:

5

= \frac{1 - 5}{10}

二次方程组的解是:

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1}{5 u}

的分母，令其等于零

解

5 u = 0 : u = 0

5 u = 0

两边除以

5

5 u = 0

两边除以

5

\frac{5 u}{5} = \frac{0}{5}

化简

u = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = \frac{1 + 5}{10}, u = \frac{1 - 5}{10}

u = tan (y)

代回

tan (y) = \frac{1 + 5}{10}, tan (y) = \frac{1 - 5}{10}

tan (y) = \frac{1 + 5}{10}, tan (y) = \frac{1 - 5}{10}

tan (y) = \frac{1 + 5}{10} : y = arctan (\frac{1 + 5}{10}) + πn

tan (y) = \frac{1 + 5}{10}

使用反三角函数性质

tan (y) = \frac{1 + 5}{10}

tan (y) = \frac{1 + 5}{10}

的通解

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

y = arctan (\frac{1 + 5}{10}) + πn

y = arctan (\frac{1 + 5}{10}) + πn

tan (y) = \frac{1 - 5}{10} : y = arctan (\frac{1 - 5}{10}) + πn

tan (y) = \frac{1 - 5}{10}

使用反三角函数性质

tan (y) = \frac{1 - 5}{10}

tan (y) = \frac{1 - 5}{10}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

y = arctan (\frac{1 - 5}{10}) + πn

y = arctan (\frac{1 - 5}{10}) + πn

合并所有解

y = arctan (\frac{1 + 5}{10}) + πn, y = arctan (\frac{1 - 5}{10}) + πn

以小数形式表示解

y = 0.31297 \dots + πn, y = - 0.12298 \dots + πn

作图

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