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受欢迎的三角函数 >

((1-tan^2(a)))/((tan(a)))=2cot^2(a)

解答

\frac{( 1 - tan ^{2} ( a ) )}{( tan ( a ) )} = 2 cot^{2} (a)

解答

a = 2.15228 \dots + πn

+1

度数

a = 123.31684 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{( 1 - tan ^{2} ( a ) )}{( tan ( a ) )} = 2 cot^{2} (a)

两边减去

2 cot^{2} (a)

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a) = 0

化简

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a) : \frac{1 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

\frac{1 - tan ^{2} ( a )}{tan ( a )} - 2 cot^{2} (a)

将项转换为分式:

2 cot^{2} (a) = \frac{2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

= \frac{1 - tan ^{2} ( a )}{tan ( a )} - \frac{2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

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 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )}

\frac{1 - tan ^{2} ( a ) - 2 cot ^{2} ( a ) tan ( a )}{tan ( a )} = 0

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

1 - tan^{2} (a) - 2 cot^{2} (a) tan (a) = 0

使用三角恒等式改写

1 - tan^{2} (a) - 2 cot^{2} (a) tan (a)

使用基本三角恒等式:

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

= 1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )}

化简

1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )} : 1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

1 - (\frac{1}{cot ( a )})^{2} - 2 cot^{2} (a) \frac{1}{cot ( a )}

(\frac{1}{cot ( a )})^{2} = \frac{1}{cot ^{2} ( a )}

(\frac{1}{cot ( a )})^{2}

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{cot ^{2} ( a )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cot ^{2} ( a )}

2 cot^{2} (a) \frac{1}{cot ( a )} = 2 cot (a)

2 cot^{2} (a) \frac{1}{cot ( a )}

分式相乘:

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

= \frac{1 \cdot 2 cot ^{2} ( a )}{cot ( a )}

数字相乘:

1 \cdot 2 = 2

= \frac{2 cot ^{2} ( a )}{cot ( a )}

约分:

cot (a)

= 2 cot (a)

= 1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

= 1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a)

1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a) = 0

用替代法求解

1 - \frac{1}{cot ^{2} ( a )} - 2 cot (a) = 0

令

: cot (a) = u

1 - \frac{1}{u ^{2}} - 2 u = 0

1 - \frac{1}{u ^{2}} - 2 u = 0 : u \approx - 0.65729 \dots

1 - \frac{1}{u ^{2}} - 2 u = 0

在两边乘以

u^{2}

1 - \frac{1}{u ^{2}} - 2 u = 0

在两边乘以

u^{2}

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - 2 u u^{2} = 0 \cdot u^{2}

化简

1 \cdot u^{2} - \frac{1}{u ^{2}} u^{2} - 2 u u^{2} = 0 \cdot u^{2}

化简

1 \cdot u^{2} : u^{2}

1 \cdot u^{2}

乘以:

1 \cdot u^{2} = u^{2}

= u^{2}

化简

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

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

分式相乘:

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

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

约分:

u^{2}

= - 1

化简

- 2 u u^{2} : - 2 u^{3}

- 2 u u^{2}

使用指数法则:

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

u u^{2} = u^{1 + 2}

= - 2 u^{1 + 2}

数字相加:

1 + 2 = 3

= - 2 u^{3}

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

u^{2} - 1 - 2 u^{3} = 0

u^{2} - 1 - 2 u^{3} = 0

u^{2} - 1 - 2 u^{3} = 0

解

u^{2} - 1 - 2 u^{3} = 0 : u \approx - 0.65729 \dots

u^{2} - 1 - 2 u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a = 0

- 2 u^{3} + u^{2} - 1 = 0

使用牛顿-拉弗森方法找到

- 2 u^{3} + u^{2} - 1 = 0

的一个解:

u \approx - 0.65729 \dots

- 2 u^{3} + u^{2} - 1 = 0

牛顿-拉弗森近似法定义

f (u) = - 2 u^{3} + u^{2} - 1

找到

f^{^{'}} (u) : - 6 u^{2} + 2 u

\frac{d}{d u} (- 2 u^{3} + u^{2} - 1)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{3}) + \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 3 u^{3 - 1}

化简

= 6 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

化简

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 6 u^{2} + 2 u - 0

化简

= - 6 u^{2} + 2 u

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.75 : Δ u_{1} = 0.25

f (u_{0}) = - 2 (- 1)^{3} + (- 1)^{2} - 1 = 2

f^{^{'}} (u_{0}) = - 6 (- 1)^{2} + 2 (- 1) = - 8

u_{1} = - 0.75

Δ u_{1} = ∣ - 0.75 - (- 1) ∣ = 0.25

Δ u_{1} = 0.25

u_{2} = - 0.66666 \dots : Δ u_{2} = 0.08333 \dots

f (u_{1}) = - 2 (- 0.75)^{3} + (- 0.75)^{2} - 1 = 0.40625

f^{^{'}} (u_{1}) = - 6 (- 0.75)^{2} + 2 (- 0.75) = - 4.875

u_{2} = - 0.66666 \dots

Δ u_{2} = ∣ - 0.66666 \dots - (- 0.75) ∣ = 0.08333 \dots

Δ u_{2} = 0.08333 \dots

u_{3} = - 0.65740 \dots : Δ u_{3} = 0.00925 \dots

f (u_{2}) = - 2 (- 0.66666 \dots)^{3} + (- 0.66666 \dots)^{2} - 1 = 0.03703 \dots

f^{^{'}} (u_{2}) = - 6 (- 0.66666 \dots)^{2} + 2 (- 0.66666 \dots) = - 4

u_{3} = - 0.65740 \dots

Δ u_{3} = ∣ - 0.65740 \dots - (- 0.66666 \dots) ∣ = 0.00925 \dots

Δ u_{3} = 0.00925 \dots

u_{4} = - 0.65729 \dots : Δ u_{4} = 0.00010 \dots

f (u_{3}) = - 2 (- 0.65740 \dots)^{3} + (- 0.65740 \dots)^{2} - 1 = 0.00042 \dots

f^{^{'}} (u_{3}) = - 6 (- 0.65740 \dots)^{2} + 2 (- 0.65740 \dots) = - 3.90792 \dots

u_{4} = - 0.65729 \dots

Δ u_{4} = ∣ - 0.65729 \dots - (- 0.65740 \dots) ∣ = 0.00010 \dots

Δ u_{4} = 0.00010 \dots

u_{5} = - 0.65729 \dots : Δ u_{5} = 1.51148 E - 8

f (u_{4}) = - 2 (- 0.65729 \dots)^{3} + (- 0.65729 \dots)^{2} - 1 = 5.90512 E - 8

f^{^{'}} (u_{4}) = - 6 (- 0.65729 \dots)^{2} + 2 (- 0.65729 \dots) = - 3.90684 \dots

u_{5} = - 0.65729 \dots

Δ u_{5} = ∣ - 0.65729 \dots - (- 0.65729 \dots) ∣ = 1.51148 E - 8

Δ u_{5} = 1.51148 E - 8

u \approx - 0.65729 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 2 u ^{3} + u ^{2} - 1}{u + 0.65729 \dots} = - 2 u^{2} + 2.31459 \dots u - 1.52137 \dots

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots \approx 0

使用牛顿-拉弗森方法找到

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots = 0

的一个解:

u \in R

无解

- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 2 u^{2} + 2.31459 \dots u - 1.52137 \dots

找到

f^{^{'}} (u) : - 4 u + 2.31459 \dots

\frac{d}{d u} (- 2 u^{2} + 2.31459 \dots u - 1.52137 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{2}) + \frac{d}{d u} (2.31459 \dots u) - \frac{d}{d u} (1.52137 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

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

化简

= 4 u

\frac{d}{d u} (2.31459 \dots u) = 2.31459 \dots

\frac{d}{d u} (2.31459 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2.31459 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 2.31459 \dots \cdot 1

化简

= 2.31459 \dots

\frac{d}{d u} (1.52137 \dots) = 0

\frac{d}{d u} (1.52137 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 4 u + 2.31459 \dots - 0

化简

= - 4 u + 2.31459 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.28397 \dots : Δ u_{1} = 0.71602 \dots

f (u_{0}) = - 2 \cdot 1^{2} + 2.31459 \dots \cdot 1 - 1.52137 \dots = - 1.20678 \dots

f^{^{'}} (u_{0}) = - 4 \cdot 1 + 2.31459 \dots = - 1.68540 \dots

u_{1} = 0.28397 \dots

Δ u_{1} = ∣ 0.28397 \dots - 1 ∣ = 0.71602 \dots

Δ u_{1} = 0.71602 \dots

u_{2} = 1.15391 \dots : Δ u_{2} = 0.86993 \dots

f (u_{1}) = - 2 \cdot 0.28397 \dots^{2} + 2.31459 \dots \cdot 0.28397 \dots - 1.52137 \dots = - 1.02537 \dots

f^{^{'}} (u_{1}) = - 4 \cdot 0.28397 \dots + 2.31459 \dots = 1.17867 \dots

u_{2} = 1.15391 \dots

Δ u_{2} = ∣ 1.15391 \dots - 0.28397 \dots ∣ = 0.86993 \dots

Δ u_{2} = 0.86993 \dots

u_{3} = 0.49614 \dots : Δ u_{3} = 0.65777 \dots

f (u_{2}) = - 2 \cdot 1.15391 \dots^{2} + 2.31459 \dots \cdot 1.15391 \dots - 1.52137 \dots = - 1.51356 \dots

f^{^{'}} (u_{2}) = - 4 \cdot 1.15391 \dots + 2.31459 \dots = - 2.30105 \dots

u_{3} = 0.49614 \dots

Δ u_{3} = ∣ 0.49614 \dots - 1.15391 \dots ∣ = 0.65777 \dots

Δ u_{3} = 0.65777 \dots

u_{4} = 3.11809 \dots : Δ u_{4} = 2.62195 \dots

f (u_{3}) = - 2 \cdot 0.49614 \dots^{2} + 2.31459 \dots \cdot 0.49614 \dots - 1.52137 \dots = - 0.86532 \dots

f^{^{'}} (u_{3}) = - 4 \cdot 0.49614 \dots + 2.31459 \dots = 0.33003 \dots

u_{4} = 3.11809 \dots

Δ u_{4} = ∣ 3.11809 \dots - 0.49614 \dots ∣ = 2.62195 \dots

Δ u_{4} = 2.62195 \dots

u_{5} = 1.76452 \dots : Δ u_{5} = 1.35357 \dots

f (u_{4}) = - 2 \cdot 3.11809 \dots^{2} + 2.31459 \dots \cdot 3.11809 \dots - 1.52137 \dots = - 13.74928 \dots

f^{^{'}} (u_{4}) = - 4 \cdot 3.11809 \dots + 2.31459 \dots = - 10.15778 \dots

u_{5} = 1.76452 \dots

Δ u_{5} = ∣ 1.76452 \dots - 3.11809 \dots ∣ = 1.35357 \dots

Δ u_{5} = 1.35357 \dots

u_{6} = 0.99203 \dots : Δ u_{6} = 0.77249 \dots

f (u_{5}) = - 2 \cdot 1.76452 \dots^{2} + 2.31459 \dots \cdot 1.76452 \dots - 1.52137 \dots = - 3.66430 \dots

f^{^{'}} (u_{5}) = - 4 \cdot 1.76452 \dots + 2.31459 \dots = - 4.74350 \dots

u_{6} = 0.99203 \dots

Δ u_{6} = ∣ 0.99203 \dots - 1.76452 \dots ∣ = 0.77249 \dots

Δ u_{6} = 0.77249 \dots

u_{7} = 0.27025 \dots : Δ u_{7} = 0.72177 \dots

f (u_{6}) = - 2 \cdot 0.99203 \dots^{2} + 2.31459 \dots \cdot 0.99203 \dots - 1.52137 \dots = - 1.19348 \dots

f^{^{'}} (u_{6}) = - 4 \cdot 0.99203 \dots + 2.31459 \dots = - 1.65353 \dots

u_{7} = 0.27025 \dots

Δ u_{7} = ∣ 0.27025 \dots - 0.99203 \dots ∣ = 0.72177 \dots

Δ u_{7} = 0.72177 \dots

u_{8} = 1.11489 \dots : Δ u_{8} = 0.84464 \dots

f (u_{7}) = - 2 \cdot 0.27025 \dots^{2} + 2.31459 \dots \cdot 0.27025 \dots - 1.52137 \dots = - 1.04192 \dots

f^{^{'}} (u_{7}) = - 4 \cdot 0.27025 \dots + 2.31459 \dots = 1.23356 \dots

u_{8} = 1.11489 \dots

Δ u_{8} = ∣ 1.11489 \dots - 0.27025 \dots ∣ = 0.84464 \dots

Δ u_{8} = 0.84464 \dots

u_{9} = 0.44970 \dots : Δ u_{9} = 0.66519 \dots

f (u_{8}) = - 2 \cdot 1.11489 \dots^{2} + 2.31459 \dots \cdot 1.11489 \dots - 1.52137 \dots = - 1.42683 \dots

f^{^{'}} (u_{8}) = - 4 \cdot 1.11489 \dots + 2.31459 \dots = - 2.14500 \dots

u_{9} = 0.44970 \dots

Δ u_{9} = ∣ 0.44970 \dots - 1.11489 \dots ∣ = 0.66519 \dots

Δ u_{9} = 0.66519 \dots

u_{10} = 2.16551 \dots : Δ u_{10} = 1.71580 \dots

f (u_{9}) = - 2 \cdot 0.44970 \dots^{2} + 2.31459 \dots \cdot 0.44970 \dots - 1.52137 \dots = - 0.88496 \dots

f^{^{'}} (u_{9}) = - 4 \cdot 0.44970 \dots + 2.31459 \dots = 0.51576 \dots

u_{10} = 2.16551 \dots

Δ u_{10} = ∣ 2.16551 \dots - 0.44970 \dots ∣ = 1.71580 \dots

Δ u_{10} = 1.71580 \dots

无法得出解

解是

u \approx - 0.65729 \dots

u \approx - 0.65729 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 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 \approx - 0.65729 \dots

u = cot (a)

代回

cot (a) \approx - 0.65729 \dots

cot (a) \approx - 0.65729 \dots

cot (a) = - 0.65729 \dots : a = arccot (- 0.65729 \dots) + πn

cot (a) = - 0.65729 \dots

使用反三角函数性质

cot (a) = - 0.65729 \dots

cot (a) = - 0.65729 \dots

的通解

cot (x) = - a \Rightarrow x = arccot (- a) + πn

a = arccot (- 0.65729 \dots) + πn

a = arccot (- 0.65729 \dots) + πn

合并所有解

a = arccot (- 0.65729 \dots) + πn

以小数形式表示解

a = 2.15228 \dots + πn

作图

流行的例子

cos(6x)=cos(2x)

cos (6 x) = cos (2 x)

cot(x)cos(x)-sin(x)=1

cot (x) cos (x) - sin (x) = 1

cos^4(x)+cos^3(x)-2=0

cos^{4} (x) + cos^{3} (x) - 2 = 0

tan^2(x)= 1/(cos(x)+1)

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

(sin^2(x)-2cos(x)+1)/4 =0

\frac{sin ^{2} ( x ) - 2 cos ( x ) + 1}{4} = 0