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受欢迎的三角函数 >

csc^2(θ)+1=cot^2(θ)

解答

csc^{2} (θ) + 1 = cot^{2} (θ)

解答

θ \in R 无解

求解步骤

csc^{2} (θ) + 1 = cot^{2} (θ)

两边减去

cot^{2} (θ)

csc^{2} (θ) + 1 - cot^{2} (θ) = 0

用 sin, cos 表示

1 - cot^{2} (θ) + csc^{2} (θ)

使用基本三角恒等式:

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

= 1 - (\frac{cos ( θ )}{sin ( θ )})^{2} + csc^{2} (θ)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

化简

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

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

(\frac{cos ( θ )}{sin ( θ )})^{2} = \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

(\frac{cos ( θ )}{sin ( θ )})^{2}

使用指数法则:

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

= \frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

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

(\frac{1}{sin ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

= \frac{1 \cdot sin ^{2} ( θ )}{sin ^{2} ( θ )} + \frac{- cos ^{2} ( θ ) + 1}{sin ^{2} ( θ )}

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 \cdot sin ^{2} ( θ ) - cos ^{2} ( θ ) + 1}{sin ^{2} ( θ )}

乘以:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

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

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

1 - cos^{2} (θ) + sin^{2} (θ) = 0

使用三角恒等式改写

1 - cos^{2} (θ) + sin^{2} (θ)

使用倍角公式:

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

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

= 1 - cos (2 θ)

1 - cos (2 θ) = 0

将

1

到右边

1 - cos (2 θ) = 0

两边减去

1

1 - cos (2 θ) - 1 = 0 - 1

化简

- cos (2 θ) = - 1

- cos (2 θ) = - 1

两边除以

- 1

- cos (2 θ) = - 1

两边除以

- 1

\frac{- cos ( 2 θ )}{- 1} = \frac{- 1}{- 1}

化简

cos (2 θ) = 1

cos (2 θ) = 1

cos (2 θ) = 1

的通解

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 θ = 0 + 2 πn

2 θ = 0 + 2 πn

解

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

两边除以

2

2 θ = 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{2 πn}{2}

化简

θ = πn

θ = πn

θ = πn

因为方程对以下值无定义:

πn

θ \in R 无解

作图

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