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受欢迎的三角函数 >

cot(x)=sin(x)

解答

cot (x) = sin (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

cot (x) = sin (x)

两边减去

sin (x)

cot (x) - sin (x) = 0

用 sin, cos 表示

cot (x) - sin (x)

使用基本三角恒等式:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

化简

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

\frac{cos ( x )}{sin ( x )} - sin (x)

将项转换为分式:

sin (x) = \frac{sin ( x ) sin ( x )}{sin ( x )}

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

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

cos (x) - sin (x) sin (x)

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

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

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

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

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

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

-sqrt(3)cos(x)-sin(x)=-sqrt(3)

- 3 cos (x) - sin (x) = - 3

9 cot^{2} (x) = 3

cos(x)=-sqrt(2)

cos (x) = - 2

tan(x)(cot(x)-1)=0

tan (x) (cot (x) - 1) = 0

sin(x)cos(x)-5cos(x)=0

sin (x) cos (x) - 5 cos (x) = 0