zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{1}{2}

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

化简

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

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

将项转换为分式:

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

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

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

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

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

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

使用指数法则:

a^{b} = a^{2} a^{b - 2}

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

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

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

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

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

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

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

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = sin^{2} (x)

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot sin^{2} (x) = 2 sin^{2} (x)

2 \cdot 1 \cdot sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

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

使用指数法则:

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

= sin^{2 \cdot 2} (x)

数字相乘:

2 \cdot 2 = 4

= sin^{4} (x)

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

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

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

乘开

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

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

打开括号

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

使用加减运算法则

+ (- a) = - a

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

化简

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

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

数字相乘:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

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

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

1 - 4 u^{2} = 0

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

1 - 4 u^{2} = 0

将

1

到右边

1 - 4 u^{2} = 0

两边减去

1

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

化简

- 4 u^{2} = - 1

- 4 u^{2} = - 1

两边除以

- 4

- 4 u^{2} = - 1

两边除以

- 4

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

化简

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

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}

\frac{1}{4} = \frac{1}{2}

\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}

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

= \frac{1}{2}

使用法则

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

化简

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

\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}

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

= \frac{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

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

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

作图

流行的例子

cos(x)= 1/4 ,0<= x<= 2pi

cos (x) = \frac{1}{4}, 0 \leq x \leq 2 π

2 = 2 cos (t)

50sin(45)=600sin(θ)

50 sin (4 5^{\circ}) = 600 sin (θ)

2+2cos(x)= 2/(1+cos(x))

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

k sin (x) - c = 0