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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

分解

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

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

将

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

改写为

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

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

使用指数法则:

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

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

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

使用指数法则:

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

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

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

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

使用平方差公式:

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

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

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

分解

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

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

使用平方差公式:

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

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

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

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

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

使用三角恒等式改写

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

= (cos (x) + sin (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{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

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

两边减去

1

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

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

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

合并所有解

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

作图

流行的例子

2 tan (x) + 4 = 0

cos (10 x) = - \frac{1}{2}

cos(2x-180)-sin(x-90)=0

cos (2 x - 18 0^{\circ}) - sin (x - 9 0^{\circ}) = 0

sin (x) tan (x) = 0

sin(2x)-sin(4x)=0

sin (2 x) - sin (4 x) = 0