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 >

sin^4(x)-cos^4(x)=0

解答

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

解答

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

求解步骤

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

分解

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

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

将

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

改写为

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

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

分解

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

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

使用平方差公式:

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

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

= (sin (x) + cos (x)) (sin (x) - cos (x))

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= (- cos (x) + sin (x)) (cos (x) + sin (x)) \cdot 1

化简

(- cos (x) + sin (x)) (cos (x) + sin (x)) \cdot 1 : (- cos (x) + sin (x)) (cos (x) + sin (x))

(- cos (x) + sin (x)) (cos (x) + sin (x)) \cdot 1

乘以:

(cos (x) + sin (x)) \cdot 1 = (cos (x) + sin (x))

= (sin (x) - cos (x)) (cos (x) + sin (x))

= (- cos (x) + sin (x)) (cos (x) + sin (x))

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

分别求解每个部分

- cos (x) + sin (x) = 0 or cos (x) + sin (x) = 0

- cos (x) + sin (x) = 0 : x = \frac{π}{4} + πn

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

使用三角恒等式改写

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

在两边除以

cos (x), cos (x) \neq = 0

\frac{- cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

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

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 1 + tan (x) = 0

- 1 + tan (x) = 0

将

1

到右边

- 1 + tan (x) = 0

两边加上

1

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

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

cos (x) + sin (x) = 0 : x = \frac{3 π}{4} + πn

cos (x) + sin (x) = 0

使用三角恒等式改写

cos (x) + sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

\frac{cos ( x ) + sin ( x )}{cos ( x )} = \frac{0}{cos ( x )}

化简

1 + \frac{sin ( x )}{cos ( x )} = 0

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (x) = 0

1 + tan (x) = 0

将

1

到右边

1 + tan (x) = 0

两边减去

1

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

化简

tan (x) = - 1

tan (x) = - 1

tan (x) = - 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

作图

流行的例子

tan(θ)cos(θ)+cos(θ)=0

tan (θ) cos (θ) + cos (θ) = 0

3 cos (θ) = 0

2sin^2(θ)+sin(θ)-9=2sin(θ)-8

2 sin^{2} (θ) + sin (θ) - 9 = 2 sin (θ) - 8

cos(θ)=(sqrt(5))/3

cos (θ) = \frac{5}{3}

-16sin(x)+8sin(2x)=0

- 16 sin (x) + 8 sin (2 x) = 0