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受欢迎的三角函数 >

1/(2cos^2(x-1))=(1+tan^2(x))/(2sec^2(x))

解答

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

解答

x = π + 2 πn + 1, x = 2 πn + 1

+1

度数

x = 237.29577 \dots^{\circ} + 36 0^{\circ} n, x = 57.29577 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

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

化简

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

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

2 cos^{2} (x - 1), 2 sec^{2} (x)

的最小公倍数:

2 cos^{2} (x - 1) sec^{2} (x)

2 cos^{2} (x - 1), 2 sec^{2} (x)

最小公倍数 (LCM)

2, 2

的最小公倍数:

2

2, 2

最小公倍数 (LCM)

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

2

质因数分解:

2

2

2

是质数，因此无法因数分解

= 2

将每个因子乘以它在

2

或

2

中出现的最多次数

= 2

数字相乘:

2 = 2

= 2

计算出由出现在

2 cos^{2} (x - 1)

或

2 sec^{2} (x)

中的因子组成的表达式

= 2 cos^{2} (x - 1) sec^{2} (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

2 cos^{2} (x - 1) sec^{2} (x)

对于

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

将分母和分子乘以

sec^{2} (x)

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

对于

\frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )} :

将分母和分子乘以

cos^{2} (x - 1)

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

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

分解

sec^{2} (x) - cos^{2} (- 1 + x) sec^{2} (x) : - sec^{2} (x) (cos (- 1 + x) + 1) (cos (- 1 + x) - 1)

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

因式分解出通项

- sec^{2} (x)

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

分解

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

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

将

1

改写为

1^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

分别求解每个部分

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

sec^{2} (x) = 0 :

无解

sec^{2} (x) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

sec (x) = 0

sec (x) \leq - 1 or sec (x) \geq 1

无解

cos (- 1 + x) + 1 = 0 : x = π + 2 πn + 1

cos (- 1 + x) + 1 = 0

将

1

到右边

cos (- 1 + x) + 1 = 0

两边减去

1

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

化简

cos (- 1 + x) = - 1

cos (- 1 + x) = - 1

cos (- 1 + 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}

- 1 + x = π + 2 πn

- 1 + x = π + 2 πn

解

- 1 + x = π + 2 πn : x = π + 2 πn + 1

- 1 + x = π + 2 πn

将

1

到右边

- 1 + x = π + 2 πn

两边加上

1

- 1 + x + 1 = π + 2 πn + 1

化简

x = π + 2 πn + 1

x = π + 2 πn + 1

x = π + 2 πn + 1

cos (- 1 + x) - 1 = 0 : x = 2 πn + 1

cos (- 1 + x) - 1 = 0

将

1

到右边

cos (- 1 + x) - 1 = 0

两边加上

1

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

化简

cos (- 1 + x) = 1

cos (- 1 + x) = 1

cos (- 1 + 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}

- 1 + x = 0 + 2 πn

- 1 + x = 0 + 2 πn

解

- 1 + x = 0 + 2 πn : x = 2 πn + 1

- 1 + x = 0 + 2 πn

0 + 2 πn = 2 πn

- 1 + x = 2 πn

将

1

到右边

- 1 + x = 2 πn

两边加上

1

- 1 + x + 1 = 2 πn + 1

化简

x = 2 πn + 1

x = 2 πn + 1

x = 2 πn + 1

合并所有解

x = π + 2 πn + 1, x = 2 πn + 1

作图

流行的例子

sin(2x)-2cos(2x)=0

sin (2 x) - 2 cos (2 x) = 0

tan(x)=(87.092)/(86.676)

tan (x) = \frac{87.092}{86.676}

csc (x) = - \frac{3}{2}

1 + 3 sin (θ) = 0

(tan(2x))/(sin(x))=-2

\frac{tan ( 2 x )}{sin ( x )} = - 2