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受欢迎的三角函数 >

cos(4y)=2cos(2y)-1

解答

cos (4 y) = 2 cos (2 y) - 1

解答

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

+1

度数

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

求解步骤

cos (4 y) = 2 cos (2 y) - 1

两边减去

2 cos (2 y) - 1

cos (4 y) - 2 cos (2 y) + 1 = 0

使用三角恒等式改写

1 + cos (4 y) - 2 cos (2 y)

cos (4 y) = 2 cos^{2} (2 y) - 1

cos (4 y)

改写为

= cos (2 \cdot 2 y)

使用倍角公式:

cos (2 y) = 2 cos^{2} (y) - 1

cos (2 \cdot 2 y) = 2 cos^{2} (2 y) - 1

= 2 cos^{2} (2 y) - 1

= 1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y)

化简

1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y) : 2 cos^{2} (2 y) - 2 cos (2 y)

1 + 2 cos^{2} (2 y) - 1 - 2 cos (2 y)

对同类项分组

= 2 cos^{2} (2 y) - 2 cos (2 y) + 1 - 1

1 - 1 = 0

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

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

- 2 cos (2 y) + 2 cos^{2} (2 y) = 0

用替代法求解

- 2 cos (2 y) + 2 cos^{2} (2 y) = 0

令

: cos (2 y) = u

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

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

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

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} - 2 u = 0

使用求根公式求解

2 u^{2} - 2 u = 0

二次方程求根公式:

若

a = 2, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 2 \cdot 0

使用指数法则:

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

若

n

是偶数

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

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

使用法则

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 2 ) - 2}{2 \cdot 2} : 0

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

使用法则

- (- a) = a

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

数字相减:

2 - 2 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

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

= 0

二次方程组的解是:

u = 1, u = 0

u = cos (2 y)

代回

cos (2 y) = 1, cos (2 y) = 0

cos (2 y) = 1, cos (2 y) = 0

cos (2 y) = 1 : y = πn

cos (2 y) = 1

cos (2 y) = 1

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}

2 y = 0 + 2 πn

2 y = 0 + 2 πn

解

2 y = 0 + 2 πn : y = πn

2 y = 0 + 2 πn

0 + 2 πn = 2 πn

2 y = 2 πn

两边除以

2

2 y = 2 πn

两边除以

2

\frac{2 y}{2} = \frac{2 πn}{2}

化简

y = πn

y = πn

y = πn

cos (2 y) = 0 : y = \frac{π}{4} + πn, y = \frac{3 π}{4} + πn

cos (2 y) = 0

cos (2 y) = 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}

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

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

解

2 y = \frac{π}{2} + 2 πn : y = \frac{π}{4} + πn

2 y = \frac{π}{2} + 2 πn

两边除以

2

2 y = \frac{π}{2} + 2 πn

两边除以

2

\frac{2 y}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 y}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 y}{2} : y

\frac{2 y}{2}

数字相除:

\frac{2}{2} = 1

= y

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

y = \frac{π}{4} + πn

y = \frac{π}{4} + πn

y = \frac{π}{4} + πn

解

2 y = \frac{3 π}{2} + 2 πn : y = \frac{3 π}{4} + πn

2 y = \frac{3 π}{2} + 2 πn

两边除以

2

2 y = \frac{3 π}{2} + 2 πn

两边除以

2

\frac{2 y}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 y}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 y}{2} : y

\frac{2 y}{2}

数字相除:

\frac{2}{2} = 1

= y

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

y = \frac{3 π}{4} + πn

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

合并所有解

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

作图

流行的例子

csc(x)cot(x)=2sqrt(3)

csc (x) cot (x) = 23

cos^2(θ)+2sin(θ)+1=0

cos^{2} (θ) + 2 sin (θ) + 1 = 0

arcsec(x)=arcsec(2)

arcsec (x) = arcsec (2)

tan^2(x)+(sqrt(3)-1)tan(x)-sqrt(3)=0

tan^{2} (x) + (3 - 1) tan (x) - 3 = 0

2 sec (x) - 5 = 0