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受欢迎的三角函数 >

csc^2(x)=sec(x)

解答

csc^{2} (x) = sec (x)

解答

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度数

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

csc^{2} (x) = sec (x)

两边减去

sec (x)

csc^{2} (x) - sec (x) = 0

用 sin, cos 表示

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

化简

(\frac{1}{sin ( x )})^{2} - \frac{1}{cos ( x )} : \frac{cos ( x ) - sin ^{2} ( x )}{sin ^{2} ( x ) cos ( x )}

(\frac{1}{sin ( x )})^{2} - \frac{1}{cos ( x )}

(\frac{1}{sin ( x )})^{2} = \frac{1}{sin ^{2} ( x )}

(\frac{1}{sin ( x )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( x )}

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

sin^{2} (x), cos (x)

的最小公倍数:

sin^{2} (x) cos (x)

sin^{2} (x), cos (x)

最小公倍数 (LCM)

计算出由出现在

sin^{2} (x)

或

cos (x)

中的因子组成的表达式

= sin^{2} (x) cos (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin^{2} (x) cos (x)

对于

\frac{1}{sin ^{2} ( x )} :

将分母和分子乘以

cos (x)

\frac{1}{sin ^{2} ( x )} = \frac{1 \cdot cos ( x )}{sin ^{2} ( x ) cos ( x )} = \frac{cos ( x )}{sin ^{2} ( x ) cos ( x )}

对于

\frac{1}{cos ( x )} :

将分母和分子乘以

sin^{2} (x)

\frac{1}{cos ( x )} = \frac{1 \cdot sin ^{2} ( x )}{cos ( x ) sin ^{2} ( x )} = \frac{sin ^{2} ( x )}{sin ^{2} ( x ) cos ( x )}

= \frac{cos ( x )}{sin ^{2} ( x ) cos ( x )} - \frac{sin ^{2} ( x )}{sin ^{2} ( x ) cos ( x )}

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{cos ( x ) - sin ^{2} ( x )}{sin ^{2} ( x ) cos ( x )}

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

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

cos (x) - sin^{2} (x) = 0

两边加上

sin^{2} (x)

cos (x) = sin^{2} (x)

两边进行平方

cos^{2} (x) = (sin^{2} (x))^{2}

两边减去

(sin^{2} (x))^{2}

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

分解

cos^{2} (x) - sin^{4} (x) : (cos (x) + sin^{2} (x)) (cos (x) - sin^{2} (x))

cos^{2} (x) - sin^{4} (x)

使用指数法则:

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

sin^{4} (x) = (sin^{2} (x))^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

cos^{2} (x) - (sin^{2} (x))^{2} = (cos (x) + sin^{2} (x)) (cos (x) - sin^{2} (x))

= (cos (x) + sin^{2} (x)) (cos (x) - sin^{2} (x))

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

分别求解每个部分

cos (x) + sin^{2} (x) = 0 or cos (x) - sin^{2} (x) = 0

cos (x) + sin^{2} (x) = 0 : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) + sin^{2} (x) = 0

使用三角恒等式改写

cos (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= cos (x) + 1 - cos^{2} (x)

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

用替代法求解

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

令

: cos (x) = u

1 + u - u^{2} = 0

1 + u - u^{2} = 0 : u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

1 + u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = 1, c = 1

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

\frac{- 1 + 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

使用分式法则:

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

= - \frac{- 1 + 5}{2}

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

\frac{- 1 - 5}{2 ( - 1 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

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

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2} : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = - \frac{- 1 + 5}{2}

使用反三角函数性质

cos (x) = - \frac{- 1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{1 + 5}{2} :

无解

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) - sin^{2} (x) = 0 : x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) - sin^{2} (x) = 0

使用三角恒等式改写

cos (x) - sin^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

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

- (1 - cos^{2} (x)) : - 1 + cos^{2} (x)

- (1 - cos^{2} (x))

打开括号

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

使用加减运算法则

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

= - 1 + cos^{2} (x)

= cos (x) - 1 + cos^{2} (x)

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

用替代法求解

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

令

: cos (x) = u

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

- 1 + u + u^{2} = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

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

改写成标准形式

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

u^{2} + u - 1 = 0

使用求根公式求解

u^{2} + u - 1 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 1

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

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

\frac{- 1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2}, cos (x) = \frac{- 1 - 5}{2}

cos (x) = \frac{- 1 + 5}{2} : x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{- 1 + 5}{2}

使用反三角函数性质

cos (x) = \frac{- 1 + 5}{2}

cos (x) = \frac{- 1 + 5}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{- 1 - 5}{2} :

无解

cos (x) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

合并所有解

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

将解代入原方程进行验证

将它们代入

csc^{2} (x) = sec (x)

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 1 + 5}{2}) + 2 πn

的解:

假

arccos (- \frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

arccos (- \frac{- 1 + 5}{2}) + 2 π 1

对于

csc^{2} (x) = sec (x)

代入

x = arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = sec (arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

整理后得

1.61803 \dots = - 1.61803 \dots

\Rightarrow 假

检验

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

的解:

假

- arccos (- \frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

- arccos (- \frac{- 1 + 5}{2}) + 2 π 1

对于

csc^{2} (x) = sec (x)

代入

x = - arccos (- \frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1) = sec (- arccos (- \frac{- 1 + 5}{2}) + 2 π 1)

整理后得

1.61803 \dots = - 1.61803 \dots

\Rightarrow 假

检验

arccos (\frac{- 1 + 5}{2}) + 2 πn

的解:

真

arccos (\frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

arccos (\frac{- 1 + 5}{2}) + 2 π 1

对于

csc^{2} (x) = sec (x)

代入

x = arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (arccos (\frac{- 1 + 5}{2}) + 2 π 1) = sec (arccos (\frac{- 1 + 5}{2}) + 2 π 1)

整理后得

1.61803 \dots = 1.61803 \dots

\Rightarrow 真

检验

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

的解:

真

2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

代入

n = 1

2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

对于

csc^{2} (x) = sec (x)

代入

x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1

csc^{2} (2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1) = sec (2 π - arccos (\frac{- 1 + 5}{2}) + 2 π 1)

整理后得

1.61803 \dots = 1.61803 \dots

\Rightarrow 真

x = arccos (\frac{- 1 + 5}{2}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

以小数形式表示解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

(2cos(x)-sin^2(x))=1+cos^2(x)

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

6cos^3(x)+cos^2(x)-1=0

6 cos^{3} (x) + cos^{2} (x) - 1 = 0

4tan^2(x)+12tan(x)-27=0

4 tan^{2} (x) + 12 tan (x) - 27 = 0

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

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

cos^4(a)=3+4cos^2(a)+cos^4(a)

cos^{4} (a) = 3 + 4 cos^{2} (a) + cos^{4} (a)