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受欢迎的三角函数 >

sec(b)cot(b)=sin(b)

解答

sec (b) cot (b) = sin (b)

解答

b \in R 无解

求解步骤

sec (b) cot (b) = sin (b)

两边减去

sin (b)

sec (b) cot (b) - sin (b) = 0

用 sin, cos 表示

- sin (b) + cot (b) sec (b)

使用基本三角恒等式:

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

= - sin (b) + \frac{cos ( b )}{sin ( b )} sec (b)

使用基本三角恒等式:

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

= - sin (b) + \frac{cos ( b )}{sin ( b )} \cdot \frac{1}{cos ( b )}

化简

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

- sin (b) + \frac{cos ( b )}{sin ( b )} \cdot \frac{1}{cos ( b )}

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{cos ( b )} = \frac{1}{sin ( b )}

\frac{cos ( b )}{sin ( b )} \cdot \frac{1}{cos ( b )}

分式相乘:

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

= \frac{cos ( b ) \cdot 1}{sin ( b ) cos ( b )}

约分:

cos (b)

= \frac{1}{sin ( b )}

= - sin (b) + \frac{1}{sin ( b )}

将项转换为分式:

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

= - \frac{sin ( b ) sin ( b )}{sin ( b )} + \frac{1}{sin ( b )}

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{- sin ( b ) sin ( b ) + 1}{sin ( b )}

- sin (b) sin (b) + 1 = - sin^{2} (b) + 1

- sin (b) sin (b) + 1

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

sin (b) sin (b)

使用指数法则:

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

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

= sin^{1 + 1} (b)

数字相加:

1 + 1 = 2

= sin^{2} (b)

= - sin^{2} (b) + 1

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

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

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

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

1 - sin^{2} (b) = 0

用替代法求解

1 - sin^{2} (b) = 0

令

: sin (b) = u

1 - u^{2} = 0

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = sin (b)

代回

sin (b) = 1, sin (b) = - 1

sin (b) = 1, sin (b) = - 1

sin (b) = 1 : b = \frac{π}{2} + 2 πn

sin (b) = 1

sin (b) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

b = \frac{π}{2} + 2 πn

b = \frac{π}{2} + 2 πn

sin (b) = - 1 : b = \frac{3 π}{2} + 2 πn

sin (b) = - 1

sin (b) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

b = \frac{3 π}{2} + 2 πn

b = \frac{3 π}{2} + 2 πn

合并所有解

b = \frac{π}{2} + 2 πn, b = \frac{3 π}{2} + 2 πn

因为方程对以下值无定义:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

b \in R 无解

作图

流行的例子

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\frac{37}{sin ( x )} = \frac{30}{sin ( 3 0 ^{\circ} )}

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

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