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受欢迎的三角函数 >

csc(x)sec(x)=tan(x)

解答

csc (x) sec (x) = tan (x)

解答

x \in R 无解

求解步骤

csc (x) sec (x) = tan (x)

两边减去

tan (x)

csc (x) sec (x) - tan (x) = 0

用 sin, cos 表示

- tan (x) + csc (x) sec (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 1 = 1

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

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

cos (x), sin (x) cos (x)

的最小公倍数:

cos (x) sin (x)

cos (x), sin (x) cos (x)

最小公倍数 (LCM)

计算出由出现在

cos (x)

或

sin (x) cos (x)

中的因子组成的表达式

= cos (x) sin (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

cos (x) sin (x)

对于

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

将分母和分子乘以

sin (x)

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

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

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

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

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

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

用替代法求解

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

令

: sin (x) = 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 (x)

代回

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

sin (x) = - 1

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \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

x \in R 无解

作图

流行的例子

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

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

sqrt(3)tan(3x)+sqrt(3)=0,0<= x<= 180

3 tan (3 x) + 3 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

2tan(θ)+4=tan(θ)+5

2 tan (θ) + 4 = tan (θ) + 5

(sin(42)}{6.25}=\frac{sin(x))/9

\frac{sin ( 4 2 ^{\circ} )}{6.25} = \frac{sin ( x )}{9}

(tan(x)+cot(x))/(csc(x))=cos(x)

\frac{tan ( x ) + cot ( x )}{csc ( x )} = cos (x)