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受欢迎的三角函数 >

tan(23θ)+cot(23θ)= 4/(sqrt(3))

解答

tan (23 θ) + cot (23 θ) = \frac{4}{3}

解答

θ = \frac{π}{138} + \frac{πn}{23}, θ = \frac{π}{69} + \frac{πn}{23}

+1

度数

θ = 1.30434 \dots^{\circ} + 7.82608 \dots^{\circ} n, θ = 2.60869 \dots^{\circ} + 7.82608 \dots^{\circ} n

求解步骤

tan (23 θ) + cot (23 θ) = \frac{4}{3}

两边减去

\frac{4}{3}

tan (23 θ) + cot (23 θ) - \frac{4}{3} = 0

化简

tan (23 θ) + cot (23 θ) - \frac{4}{3} : \frac{3 tan ( 23 θ ) + 3 cot ( 23 θ ) - 4}{3}

tan (23 θ) + cot (23 θ) - \frac{4}{3}

将项转换为分式:

tan (23 θ) = \frac{tan ( 23 θ ) 3}{3}, cot (23 θ) = \frac{cot ( 23 θ ) 3}{3}

= \frac{tan ( 23 θ ) 3}{3} + \frac{cot ( 23 θ ) 3}{3} - \frac{4}{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{tan ( 23 θ ) 3 + cot ( 23 θ ) 3 - 4}{3}

\frac{3 tan ( 23 θ ) + 3 cot ( 23 θ ) - 4}{3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 tan (23 θ) + 3 cot (23 θ) - 4 = 0

使用三角恒等式改写

- 4 + cot (23 θ) 3 + 3 tan (23 θ)

使用基本三角恒等式:

tan (x) = \frac{1}{cot ( x )}

= - 4 + cot (23 θ) 3 + 3 \frac{1}{cot ( 23 θ )}

3 \frac{1}{cot ( 23 θ )} = \frac{3}{cot ( 23 θ )}

3 \frac{1}{cot ( 23 θ )}

分式相乘:

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

= \frac{1 \cdot 3}{cot ( 23 θ )}

乘以:

1 \cdot 3 = 3

= \frac{3}{cot ( 23 θ )}

= - 4 + 3 cot (23 θ) + \frac{3}{cot ( 23 θ )}

- 4 + \frac{3}{cot ( 23 θ )} + cot (23 θ) 3 = 0

用替代法求解

- 4 + \frac{3}{cot ( 23 θ )} + cot (23 θ) 3 = 0

令

: cot (23 θ) = u

- 4 + \frac{3}{u} + u 3 = 0

- 4 + \frac{3}{u} + u 3 = 0 : u = 3, u = \frac{3}{3}

- 4 + \frac{3}{u} + u 3 = 0

在两边乘以

u

- 4 + \frac{3}{u} + u 3 = 0

在两边乘以

u

- 4 u + \frac{3}{u} u + u 3 u = 0 \cdot u

化简

- 4 u + \frac{3}{u} u + u 3 u = 0 \cdot u

化简

\frac{3}{u} u : 3

\frac{3}{u} u

分式相乘:

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

= \frac{3 u}{u}

约分:

u

= 3

化简

u 3 u : 3 u^{2}

u 3 u

使用指数法则:

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

uu = u^{1 + 1}

= 3 u^{1 + 1}

数字相加:

1 + 1 = 2

= 3 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

- 4 u + 3 + 3 u^{2} = 0

- 4 u + 3 + 3 u^{2} = 0

- 4 u + 3 + 3 u^{2} = 0

解

- 4 u + 3 + 3 u^{2} = 0 : u = 3, u = \frac{3}{3}

- 4 u + 3 + 3 u^{2} = 0

改写成标准形式

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

3 u^{2} - 4 u + 3 = 0

使用求根公式求解

3 u^{2} - 4 u + 3 = 0

二次方程求根公式:

若

a = 3, b = - 4, c = 3

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 3 3}{2 3}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 3 3}{2 3}

(- 4)^{2} - 433 = 2

(- 4)^{2} - 433

(- 4)^{2} = 4^{2}

(- 4)^{2}

使用指数法则:

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

若

n

是偶数

(- 4)^{2} = 4^{2}

= 4^{2}

433 = 12

433

使用根式运算法则:

a a = a

33 = 3

= 4 \cdot 3

数字相乘:

4 \cdot 3 = 12

= 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数字相减:

16 - 12 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 4 ) \pm 2}{2 3}

将解分隔开

u_{1} = \frac{- ( - 4 ) + 2}{2 3}, u_{2} = \frac{- ( - 4 ) - 2}{2 3}

u = \frac{- ( - 4 ) + 2}{2 3} : 3

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

使用法则

- (- a) = a

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

数字相加:

4 + 2 = 6

= \frac{6}{2 3}

数字相除:

\frac{6}{2} = 3

= \frac{3}{3}

使用根式运算法则:

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

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

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

使用指数法则:

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

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

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

数字相减:

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

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

使用根式运算法则:

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

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

= 3

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

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

使用法则

- (- a) = a

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

数字相减:

4 - 2 = 2

= \frac{2}{2 3}

数字相除:

\frac{2}{2} = 1

= \frac{1}{3}

\frac{1}{3}

有理化:

\frac{3}{3}

\frac{1}{3}

乘以共轭根式

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3}{3}

= \frac{3}{3}

二次方程组的解是:

u = 3, u = \frac{3}{3}

u = 3, u = \frac{3}{3}

验证解

找到无定义的点（奇点）:

u = 0

取

- 4 + \frac{3}{u} + u 3

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 3, u = \frac{3}{3}

u = cot (23 θ)

代回

cot (23 θ) = 3, cot (23 θ) = \frac{3}{3}

cot (23 θ) = 3, cot (23 θ) = \frac{3}{3}

cot (23 θ) = 3 : θ = \frac{π}{138} + \frac{πn}{23}

cot (23 θ) = 3

cot (23 θ) = 3

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

23 θ = \frac{π}{6} + πn

23 θ = \frac{π}{6} + πn

解

23 θ = \frac{π}{6} + πn : θ = \frac{π}{138} + \frac{πn}{23}

23 θ = \frac{π}{6} + πn

两边除以

23

23 θ = \frac{π}{6} + πn

两边除以

23

\frac{23 θ}{23} = \frac{\frac{π}{6}}{23} + \frac{πn}{23}

化简

\frac{23 θ}{23} = \frac{\frac{π}{6}}{23} + \frac{πn}{23}

化简

\frac{23 θ}{23} : θ

\frac{23 θ}{23}

数字相除:

\frac{23}{23} = 1

= θ

化简

\frac{\frac{π}{6}}{23} + \frac{πn}{23} : \frac{π}{138} + \frac{πn}{23}

\frac{\frac{π}{6}}{23} + \frac{πn}{23}

\frac{\frac{π}{6}}{23} = \frac{π}{138}

\frac{\frac{π}{6}}{23}

使用分式法则:

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

= \frac{π}{6 \cdot 23}

数字相乘:

6 \cdot 23 = 138

= \frac{π}{138}

= \frac{π}{138} + \frac{πn}{23}

θ = \frac{π}{138} + \frac{πn}{23}

θ = \frac{π}{138} + \frac{πn}{23}

θ = \frac{π}{138} + \frac{πn}{23}

θ = \frac{π}{138} + \frac{πn}{23}

cot (23 θ) = \frac{3}{3} : θ = \frac{π}{69} + \frac{πn}{23}

cot (23 θ) = \frac{3}{3}

cot (23 θ) = \frac{3}{3}

的通解

cot (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

23 θ = \frac{π}{3} + πn

23 θ = \frac{π}{3} + πn

解

23 θ = \frac{π}{3} + πn : θ = \frac{π}{69} + \frac{πn}{23}

23 θ = \frac{π}{3} + πn

两边除以

23

23 θ = \frac{π}{3} + πn

两边除以

23

\frac{23 θ}{23} = \frac{\frac{π}{3}}{23} + \frac{πn}{23}

化简

\frac{23 θ}{23} = \frac{\frac{π}{3}}{23} + \frac{πn}{23}

化简

\frac{23 θ}{23} : θ

\frac{23 θ}{23}

数字相除:

\frac{23}{23} = 1

= θ

化简

\frac{\frac{π}{3}}{23} + \frac{πn}{23} : \frac{π}{69} + \frac{πn}{23}

\frac{\frac{π}{3}}{23} + \frac{πn}{23}

\frac{\frac{π}{3}}{23} = \frac{π}{69}

\frac{\frac{π}{3}}{23}

使用分式法则:

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

= \frac{π}{3 \cdot 23}

数字相乘:

3 \cdot 23 = 69

= \frac{π}{69}

= \frac{π}{69} + \frac{πn}{23}

θ = \frac{π}{69} + \frac{πn}{23}

θ = \frac{π}{69} + \frac{πn}{23}

θ = \frac{π}{69} + \frac{πn}{23}

θ = \frac{π}{69} + \frac{πn}{23}

合并所有解

θ = \frac{π}{138} + \frac{πn}{23}, θ = \frac{π}{69} + \frac{πn}{23}

作图

流行的例子

sin (θ) = 0.9903

tan (θ) = \frac{17}{7}

solvefor θ,cos(θ)= 6/(sqrt(85))

so l v e f or θ, cos (θ) = \frac{6}{85}

6 sin^{2} (x) - 3 = 0

2cos^2(θ)=sin(θ)+1

2 cos^{2} (θ) = sin (θ) + 1