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受欢迎的三角函数 >

1+sec^2(x)=tan^2(x)

解答

1 + sec^{2} (x) = tan^{2} (x)

解答

x \in R 无解

求解步骤

1 + sec^{2} (x) = tan^{2} (x)

两边减去

tan^{2} (x)

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

用 sin, cos 表示

1 + sec^{2} (x) - tan^{2} (x)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用指数法则:

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

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

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

合并分式

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

使用法则

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

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

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

将项转换为分式:

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

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

乘以:

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

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

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

= 1 + cos (2 x)

1 + cos (2 x) = 0

将

1

到右边

1 + cos (2 x) = 0

两边减去

1

1 + cos (2 x) - 1 = 0 - 1

化简

cos (2 x) = - 1

cos (2 x) = - 1

cos (2 x) = - 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 x = π + 2 πn

2 x = π + 2 πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

因为方程对以下值无定义:

\frac{π}{2} + πn

x \in R 无解

作图

流行的例子

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cos (t) = \frac{1}{2}

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6 tan (x) = 18 cot (x)

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