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受欢迎的三角函数 >

cos(2x)sin(x)=1

解答

cos (2 x) sin (x) = 1

解答

x = \frac{3 π}{2} + 2 πn

+1

度数

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

求解步骤

cos (2 x) sin (x) = 1

两边减去

1

cos (2 x) sin (x) - 1 = 0

使用三角恒等式改写

- 1 + cos (2 x) sin (x)

使用倍角公式:

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

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

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

用替代法求解

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

令

: sin (x) = u

- 1 + (1 - 2 u^{2}) u = 0

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

- 1 + (1 - 2 u^{2}) u = 0

展开

- 1 + (1 - 2 u^{2}) u : - 1 + u - 2 u^{3}

- 1 + (1 - 2 u^{2}) u

= - 1 + u (1 - 2 u^{2})

乘开

u (1 - 2 u^{2}) : u - 2 u^{3}

u (1 - 2 u^{2})

使用分配律:

a (b - c) = ab - a c

a = u, b = 1, c = 2 u^{2}

= u \cdot 1 - u \cdot 2 u^{2}

= 1 \cdot u - 2 u^{2} u

化简

1 \cdot u - 2 u^{2} u : u - 2 u^{3}

1 \cdot u - 2 u^{2} u

1 \cdot u = u

1 \cdot u

乘以:

1 \cdot u = u

= u

2 u^{2} u = 2 u^{3}

2 u^{2} u

使用指数法则:

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

u^{2} u = u^{2 + 1}

= 2 u^{2 + 1}

数字相加:

2 + 1 = 3

= 2 u^{3}

= u - 2 u^{3}

= u - 2 u^{3}

= - 1 + u - 2 u^{3}

- 1 + u - 2 u^{3} = 0

改写成标准形式

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

- 2 u^{3} + u - 1 = 0

因式分解

- 2 u^{3} + u - 1 : - (u + 1) (2 u^{2} - 2 u + 1)

- 2 u^{3} + u - 1

因式分解出通项

- 1

= - (2 u^{3} - u + 1)

分解

2 u^{3} - u + 1 : (u + 1) (2 u^{2} - 2 u + 1)

2 u^{3} - u + 1

使用有理根定理

a_{0} = 1, a_{n} = 2

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2

因此，检验以下有理数:

\pm \frac{1}{1 , 2}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{2 u ^{3} - u + 1}{u + 1}

\frac{2 u ^{3} - u + 1}{u + 1} = 2 u^{2} - 2 u + 1

\frac{2 u ^{3} - u + 1}{u + 1}

对

\frac{2 u ^{3} - u + 1}{u + 1}

做除法:

\frac{2 u ^{3} - u + 1}{u + 1} = 2 u^{2} + \frac{- 2 u ^{2} - u + 1}{u + 1}

将分子

2 u^{3} - u + 1

与除数

u + 1

的首项系数相除：

\frac{2 u ^{3}}{u} = 2 u^{2}

商 = 2 u^{2}

将

u + 1

乘以

2 u^{2} : 2 u^{3} + 2 u^{2}

将

2 u^{3} - u + 1

减去

2 u^{3} + 2 u^{2}

得到新的余数

余数 = - 2 u^{2} - u + 1

因此

\frac{2 u ^{3} - u + 1}{u + 1} = 2 u^{2} + \frac{- 2 u ^{2} - u + 1}{u + 1}

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

对

\frac{- 2 u ^{2} - u + 1}{u + 1}

做除法:

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

将分子

- 2 u^{2} - u + 1

与除数

u + 1

的首项系数相除：

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

商 = - 2 u

将

u + 1

乘以

- 2 u : - 2 u^{2} - 2 u

将

- 2 u^{2} - u + 1

减去

- 2 u^{2} - 2 u

得到新的余数

余数 = u + 1

因此

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

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

对

\frac{u + 1}{u + 1}

做除法:

\frac{u + 1}{u + 1} = 1

将分子

u + 1

与除数

u + 1

的首项系数相除：

\frac{u}{u} = 1

商 = 1

将

u + 1

乘以

1 : u + 1

将

u + 1

减去

u + 1

得到新的余数

余数 = 0

因此

\frac{u + 1}{u + 1} = 1

= 2 u^{2} - 2 u + 1

= 2 u^{2} - 2 u + 1

= (u + 1) (2 u^{2} - 2 u + 1)

= - (u + 1) (2 u^{2} - 2 u + 1)

- (u + 1) (2 u^{2} - 2 u + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u + 1 = 0 or 2 u^{2} - 2 u + 1 = 0

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = - 2, c = 1

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

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

化简

(- 2)^{2} - 4 \cdot 2 \cdot 1 : 2 i

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

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 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} - 8

使用虚数运算法则:

- a = i a

= i 8 - 2^{2}

- 2^{2} + 8 = 2

- 2^{2} + 8

2^{2} = 4

= - 4 + 8

数字相加/相减:

- 4 + 8 = 4

= 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

= 2 i

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 2 i}{4}

分解

2 + 2 i : 2 (1 + i)

2 + 2 i

改写为

= 2 \cdot 1 + 2 i

因式分解出通项

2

= 2 (1 + i)

= \frac{2 ( 1 + i )}{4}

约分:

2

= \frac{1 + i}{2}

将

\frac{1 + i}{2}

改写成标准复数形式:

\frac{1}{2} + \frac{1}{2} i

\frac{1 + i}{2}

使用分式法则:

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

\frac{1 + i}{2} = \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{i}{2}

= \frac{1}{2} + \frac{1}{2} i

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

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

分解

2 - 2 i : 2 (1 - i)

2 - 2 i

改写为

= 2 \cdot 1 - 2 i

因式分解出通项

2

= 2 (1 - i)

= \frac{2 ( 1 - i )}{4}

约分:

2

= \frac{1 - i}{2}

将

\frac{1 - i}{2}

改写成标准复数形式:

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

\frac{1 - i}{2}

使用分式法则:

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

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

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

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

二次方程组的解是:

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

解为

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

u = sin (x)

代回

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

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

无解

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

无解

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

无解

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

无解

合并所有解

x = \frac{3 π}{2} + 2 πn

作图

流行的例子

sin (θ) = 10

17.6^2=15^2+13.1^2-2(15)(13.1)*cos(A)

17. 6^{2} = 1 5^{2} + 13. 1^{2} - 2 (15) (13.1) \cdot cos (A)

sin (θ) = 45

cot(θ)+2csc(θ)=6

cot (θ) + 2 csc (θ) = 6

(1+cot(x))/(1+tan(x))=5

\frac{1 + cot ( x )}{1 + tan ( x )} = 5