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受欢迎的三角函数 >

cos^4(x)=(sin^2(x)-1)/4

解答

cos^{4} (x) = \frac{sin ^{2} ( x ) - 1}{4}

解答

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

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

cos^{4} (x) = \frac{sin ^{2} ( x ) - 1}{4}

两边减去

\frac{sin ^{2} ( x ) - 1}{4}

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

化简

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

cos^{4} (x) - \frac{sin ^{2} ( x ) - 1}{4}

将项转换为分式:

cos^{4} (x) = \frac{cos ^{4} ( x ) 4}{4}

= \frac{cos ^{4} ( x ) \cdot 4}{4} - \frac{sin ^{2} ( x ) - 1}{4}

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{cos ^{4} ( x ) \cdot 4 - ( sin ^{2} ( x ) - 1 )}{4}

乘开

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

cos^{4} (x) \cdot 4 - (sin^{2} (x) - 1)

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

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

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

打开括号

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

使用加减运算法则

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

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

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

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

\frac{4 cos ^{4} ( x ) - sin ^{2} ( x ) + 1}{4} = 0

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

cos^{2} (x) + 4 cos^{4} (x) = 0

用替代法求解

cos^{2} (x) + 4 cos^{4} (x) = 0

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

4 v^{2} + v = 0

解

4 v^{2} + v = 0 : v = 0, v = - \frac{1}{4}

4 v^{2} + v = 0

使用求根公式求解

4 v^{2} + v = 0

二次方程求根公式:

若

a = 4, b = 1, c = 0

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

1^{2} - 4 \cdot 4 \cdot 0 = 1

1^{2} - 4 \cdot 4 \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

v_{1} = \frac{- 1 + 1}{2 \cdot 4}, v_{2} = \frac{- 1 - 1}{2 \cdot 4}

v = \frac{- 1 + 1}{2 \cdot 4} : 0

\frac{- 1 + 1}{2 \cdot 4}

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{0}{8}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

v = \frac{- 1 - 1}{2 \cdot 4} : - \frac{1}{4}

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

数字相减:

- 1 - 1 = - 2

= \frac{- 2}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 2}{8}

使用分式法则:

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

= - \frac{2}{8}

约分:

2

= - \frac{1}{4}

二次方程组的解是:

v = 0, v = - \frac{1}{4}

v = 0, v = - \frac{1}{4}

代回

v = u^{2}

，求解

u

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

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

u = 0

解

u^{2} = - \frac{1}{4} : u = i \frac{1}{2}, u = - i \frac{1}{2}

u^{2} = - \frac{1}{4}

对于

x^{2} = f (a)

解为

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

u = - \frac{1}{4}, u = - - \frac{1}{4}

化简

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

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

= i \frac{1}{2}

使用法则

1 = 1

= i \frac{1}{2}

将

i \frac{1}{2}

改写成标准复数形式:

\frac{1}{2} i

i \frac{1}{2}

分式相乘:

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

= \frac{1 i}{2}

乘以:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

化简

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

化简

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

使用根式运算法则:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

使用虚数运算法则:

- 1 = i

= i \frac{1}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

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

= i \frac{1}{2}

= - i \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2} i

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

解为

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

u = cos (x)

代回

cos (x) = 0, cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = 0, cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}

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

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

cos (x) = i \frac{1}{2} :

无解

cos (x) = i \frac{1}{2}

无解

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

无解

cos (x) = - i \frac{1}{2}

无解

合并所有解

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

作图

流行的例子

sqrt(2)*sin(x)+1=0

2 \cdot sin (x) + 1 = 0

5cos^2(2x)+4cos^2(x)-5=0

5 cos^{2} (2 x) + 4 cos^{2} (x) - 5 = 0

sin(3x+10)=cos(x+24)

sin (3 x + 10) = cos (x + 2 4^{\circ})

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

((1+cos^2(a)))/(sin^2(a))= 5/3

\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}