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受欢迎的三角函数 >

sin^2(x)-cos^2(x)=cos^4(x)

解答

sin^{2} (x) - cos^{2} (x) = cos^{4} (x)

解答

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

+1

度数

x = 49.93964 \dots^{\circ} + 36 0^{\circ} n, x = 310.06035 \dots^{\circ} + 36 0^{\circ} n, x = 130.06035 \dots^{\circ} + 36 0^{\circ} n, x = - 130.06035 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) - cos^{2} (x) = cos^{4} (x)

两边减去

cos^{4} (x)

sin^{2} (x) - cos^{2} (x) - cos^{4} (x) = 0

使用三角恒等式改写

- cos^{2} (x) - cos^{4} (x) + sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x)

化简

- cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x) : - 2 cos^{2} (x) - cos^{4} (x) + 1

- cos^{2} (x) - cos^{4} (x) + 1 - cos^{2} (x)

对同类项分组

= - cos^{2} (x) - cos^{4} (x) - cos^{2} (x) + 1

同类项相加:

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

= - 2 cos^{2} (x) - cos^{4} (x) + 1

= - 2 cos^{2} (x) - cos^{4} (x) + 1

1 - cos^{4} (x) - 2 cos^{2} (x) = 0

用替代法求解

1 - cos^{4} (x) - 2 cos^{2} (x) = 0

令

: cos (x) = u

1 - u^{4} - 2 u^{2} = 0

1 - u^{4} - 2 u^{2} = 0 : u = i 1 + 2, u = - i 1 + 2, u = 2 - 1, u = - 2 - 1

1 - u^{4} - 2 u^{2} = 0

改写成标准形式

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

- u^{4} - 2 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

- v^{2} - 2 v + 1 = 0

解

- v^{2} - 2 v + 1 = 0 : v = - 1 - 2, v = 2 - 1

- v^{2} - 2 v + 1 = 0

使用求根公式求解

- v^{2} - 2 v + 1 = 0

二次方程求根公式:

若

a = - 1, b = - 2, c = 1

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 1 = 22

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

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 2)^{2} = 2^{2}

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

v_{1, 2} = \frac{- ( - 2 ) \pm 2 2}{2 ( - 1 )}

将解分隔开

v_{1} = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )}, v_{2} = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

v = \frac{- ( - 2 ) + 2 2}{2 ( - 1 )} : - 1 - 2

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

消掉

\frac{2 + 2 2}{2} : 1 + 2

\frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

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

数字相除:

\frac{2}{2} = 1

= 1 + 2

= - (1 + 2)

打开括号

= - (1) - (2)

使用加减运算法则

+ (- a) = - a

= - 1 - 2

v = \frac{- ( - 2 ) - 2 2}{2 ( - 1 )} : 2 - 1

\frac{- ( - 2 ) - 2 2}{2 ( - 1 )}

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

2 - 22 = - (22 - 2)

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

分解

22 - 2 : 2 (2 - 1)

22 - 2

改写为

= 22 - 2 \cdot 1

因式分解出通项

2

= 2 (2 - 1)

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

数字相除:

\frac{2}{2} = 1

= 2 - 1

二次方程组的解是:

v = - 1 - 2, v = 2 - 1

v = - 1 - 2, v = 2 - 1

代回

v = u^{2}

，求解

u

解

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

u^{2} = - 1 - 2

对于

x^{2} = f (a)

解为

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

u = - 1 - 2, u = - - 1 - 2

化简

- 1 - 2 : i 1 + 2

- 1 - 2

使用虚数运算法则:

- a = i a

= i 1 + 2

化简

- - 1 - 2 : - i 1 + 2

- - 1 - 2

使用虚数运算法则:

- a = i a

= - i 1 + 2

u = i 1 + 2, u = - i 1 + 2

解

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

u^{2} = 2 - 1

对于

x^{2} = f (a)

解为

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

u = 2 - 1, u = - 2 - 1

解为

u = i 1 + 2, u = - i 1 + 2, u = 2 - 1, u = - 2 - 1

u = cos (x)

代回

cos (x) = i 1 + 2, cos (x) = - i 1 + 2, cos (x) = 2 - 1, cos (x) = - 2 - 1

cos (x) = i 1 + 2, cos (x) = - i 1 + 2, cos (x) = 2 - 1, cos (x) = - 2 - 1

cos (x) = i 1 + 2 :

无解

cos (x) = i 1 + 2

无解

cos (x) = - i 1 + 2 :

无解

cos (x) = - i 1 + 2

无解

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

cos (x) = 2 - 1

使用反三角函数性质

cos (x) = 2 - 1

cos (x) = 2 - 1

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn

cos (x) = - 2 - 1 : x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

cos (x) = - 2 - 1

使用反三角函数性质

cos (x) = - 2 - 1

cos (x) = - 2 - 1

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

合并所有解

x = arccos (2 - 1) + 2 πn, x = 2 π - arccos (2 - 1) + 2 πn, x = arccos (- 2 - 1) + 2 πn, x = - arccos (- 2 - 1) + 2 πn

以小数形式表示解

x = 0.87161 \dots + 2 πn, x = 2 π - 0.87161 \dots + 2 πn, x = 2.26998 \dots + 2 πn, x = - 2.26998 \dots + 2 πn

作图

流行的例子

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