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受欢迎的三角函数 >

2cos^2(θ)-1=sec(θ)

解答

2 cos^{2} (θ) - 1 = sec (θ)

解答

θ = 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (θ) - 1 = sec (θ)

两边减去

sec (θ)

2 cos^{2} (θ) - 1 - sec (θ) = 0

使用三角恒等式改写

- 1 - sec (θ) + 2 cos^{2} (θ)

使用基本三角恒等式:

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

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

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

2 (\frac{1}{sec ( θ )})^{2}

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

(\frac{1}{sec ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sec ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

= 2 \cdot \frac{1}{sec ^{2} ( θ )}

分式相乘:

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

= \frac{1 \cdot 2}{sec ^{2} ( θ )}

数字相乘:

1 \cdot 2 = 2

= \frac{2}{sec ^{2} ( θ )}

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

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

用替代法求解

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

令

: sec (θ) = u

- 1 + \frac{2}{u ^{2}} - u = 0

- 1 + \frac{2}{u ^{2}} - u = 0 : u = 1, u = - 1 + i, u = - 1 - i

- 1 + \frac{2}{u ^{2}} - u = 0

在两边乘以

u^{2}

- 1 + \frac{2}{u ^{2}} - u = 0

在两边乘以

u^{2}

- 1 \cdot u^{2} + \frac{2}{u ^{2}} u^{2} - u u^{2} = 0 \cdot u^{2}

化简

- 1 \cdot u^{2} + \frac{2}{u ^{2}} u^{2} - u u^{2} = 0 \cdot u^{2}

化简

- 1 \cdot u^{2} : - u^{2}

- 1 \cdot u^{2}

乘以:

1 \cdot u^{2} = u^{2}

= - u^{2}

化简

\frac{2}{u ^{2}} u^{2} : 2

\frac{2}{u ^{2}} u^{2}

分式相乘:

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

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

约分:

u^{2}

= 2

化简

- u u^{2} : - u^{3}

- u u^{2}

使用指数法则:

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

u u^{2} = u^{1 + 2}

= - u^{1 + 2}

数字相加:

1 + 2 = 3

= - u^{3}

化简

0 \cdot u^{2} : 0

0 \cdot u^{2}

使用法则

0 \cdot a = 0

= 0

- u^{2} + 2 - u^{3} = 0

- u^{2} + 2 - u^{3} = 0

- u^{2} + 2 - u^{3} = 0

解

- u^{2} + 2 - u^{3} = 0 : u = 1, u = - 1 + i, u = - 1 - i

- u^{2} + 2 - u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{3} - u^{2} + 2 = 0

因式分解

- u^{3} - u^{2} + 2 : - (u - 1) (u^{2} + 2 u + 2)

- u^{3} - u^{2} + 2

因式分解出通项

- 1

= - (u^{3} + u^{2} - 2)

分解

u^{3} + u^{2} - 2 : (u - 1) (u^{2} + 2 u + 2)

u^{3} + u^{2} - 2

使用有理根定理

a_{0} = 2, a_{n} = 1

a_{0}

的除数

: 1, 2, a_{n}

的除数

: 1

因此，检验以下有理数:

\pm \frac{1 , 2}{1}

\frac{1}{1}

是表达式的根，所以因式分解

u - 1

= (u - 1) \frac{u ^{3} + u ^{2} - 2}{u - 1}

\frac{u ^{3} + u ^{2} - 2}{u - 1} = u^{2} + 2 u + 2

\frac{u ^{3} + u ^{2} - 2}{u - 1}

对

\frac{u ^{3} + u ^{2} - 2}{u - 1}

做除法:

\frac{u ^{3} + u ^{2} - 2}{u - 1} = u^{2} + \frac{2 u ^{2} - 2}{u - 1}

将分子

u^{3} + u^{2} - 2

与除数

u - 1

的首项系数相除：

\frac{u ^{3}}{u} = u^{2}

商 = u^{2}

将

u - 1

乘以

u^{2} : u^{3} - u^{2}

将

u^{3} + u^{2} - 2

减去

u^{3} - u^{2}

得到新的余数

余数 = 2 u^{2} - 2

因此

\frac{u ^{3} + u ^{2} - 2}{u - 1} = u^{2} + \frac{2 u ^{2} - 2}{u - 1}

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

对

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

做除法:

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

将分子

2 u^{2} - 2

与除数

u - 1

的首项系数相除：

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

商 = 2 u

将

u - 1

乘以

2 u : 2 u^{2} - 2 u

将

2 u^{2} - 2

减去

2 u^{2} - 2 u

得到新的余数

余数 = 2 u - 2

因此

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

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

对

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

做除法:

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

将分子

2 u - 2

与除数

u - 1

的首项系数相除：

\frac{2 u}{u} = 2

商 = 2

将

u - 1

乘以

2 : 2 u - 2

将

2 u - 2

减去

2 u - 2

得到新的余数

余数 = 0

因此

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

= u^{2} + 2 u + 2

= u^{2} + 2 u + 2

= (u - 1) (u^{2} + 2 u + 2)

= - (u - 1) (u^{2} + 2 u + 2)

- (u - 1) (u^{2} + 2 u + 2) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

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

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

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

u^{2} + 2 u + 2 = 0

使用求根公式求解

u^{2} + 2 u + 2 = 0

二次方程求根公式:

若

a = 1, b = 2, c = 2

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

化简

2^{2} - 4 \cdot 1 \cdot 2 : 2 i

2^{2} - 4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 2^{2} - 8

使用虚数运算法则:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

数字相加/相减:

- 4 + 8 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 i

u_{1, 2} = \frac{- 2 \pm 2 i}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 2 + 2 i}{2 \cdot 1}, u_{2} = \frac{- 2 - 2 i}{2 \cdot 1}

u = \frac{- 2 + 2 i}{2 \cdot 1} : - 1 + i

\frac{- 2 + 2 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 + 2 i}{2}

分解

- 2 + 2 i : 2 (- 1 + i)

- 2 + 2 i

改写为

= - 2 \cdot 1 + 2 i

因式分解出通项

2

= 2 (- 1 + i)

= \frac{2 ( - 1 + i )}{2}

数字相除:

\frac{2}{2} = 1

= - 1 + i

u = \frac{- 2 - 2 i}{2 \cdot 1} : - 1 - i

\frac{- 2 - 2 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 2 - 2 i}{2}

分解

- 2 - 2 i : - 2 (1 + i)

- 2 - 2 i

改写为

= - 2 \cdot 1 - 2 i

因式分解出通项

2

= - 2 (1 + i)

= - \frac{2 ( 1 + i )}{2}

数字相除:

\frac{2}{2} = 1

= - (1 + i)

取负

- (1 + i) = - 1 - i

= - 1 - i

二次方程组的解是:

u = - 1 + i, u = - 1 - i

解为

u = 1, u = - 1 + i, u = - 1 - i

u = 1, u = - 1 + i, u = - 1 - i

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = - 1 + i, u = - 1 - i

u = sec (θ)

代回

sec (θ) = 1, sec (θ) = - 1 + i, sec (θ) = - 1 - i

sec (θ) = 1, sec (θ) = - 1 + i, sec (θ) = - 1 - i

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

sec (θ) = 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sec (θ) = - 1 + i :

无解

sec (θ) = - 1 + i

无解

sec (θ) = - 1 - i :

无解

sec (θ) = - 1 - i

无解

合并所有解

θ = 2 πn

作图

流行的例子

1/(sin(x))-sin(x)=sin(x)

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

4 cos^{2} (x) = 0

sin(2x)=(2*10*1500000)/(11000000)

sin (2 x) = \frac{2 \cdot 10 \cdot 1500000}{11000000}

2/(tan(x))=3-tan(x)

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

tan (x) = 0.158