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受欢迎的三角函数 >

cos(1/(4θ))=(-sqrt(2))/2

解答

cos (\frac{1}{4 θ}) = \frac{- 2}{2}

解答

θ = \frac{1}{π ( 3 + 8 n )}, θ = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{3}{8}, n \neq = - \frac{5}{8}

+1

度数

θ = 0^{\circ} + 1.65798 \dots^{\circ} n, θ = 0^{\circ} + 1.40290 \dots^{\circ} n

求解步骤

cos (\frac{1}{4 θ}) = \frac{- 2}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

cos (\frac{1}{4 θ}) = - \frac{2}{2}

cos (\frac{1}{4 θ}) = - \frac{2}{2}

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}

\frac{1}{4 θ} = \frac{3 π}{4} + 2 πn, \frac{1}{4 θ} = \frac{5 π}{4} + 2 πn

\frac{1}{4 θ} = \frac{3 π}{4} + 2 πn, \frac{1}{4 θ} = \frac{5 π}{4} + 2 πn

解

\frac{1}{4 θ} = \frac{3 π}{4} + 2 πn : θ = \frac{1}{π ( 3 + 8 n )}; n \neq = - \frac{3}{8}

\frac{1}{4 θ} = \frac{3 π}{4} + 2 πn

在两边乘以

θ

\frac{1}{4 θ} = \frac{3 π}{4} + 2 πn

在两边乘以

θ

\frac{1}{4 θ} θ = \frac{3 π}{4} θ + 2 πn θ

化简

\frac{1}{4} = \frac{3 π}{4} θ + 2 πn θ

\frac{1}{4} = \frac{3 π}{4} θ + 2 πn θ

交换两边

\frac{3 π}{4} θ + 2 πn θ = \frac{1}{4}

在两边乘以

4

\frac{3 π}{4} θ + 2 πn θ = \frac{1}{4}

在两边乘以

4

\frac{3 π}{4} θ \cdot 4 + 2 πn θ \cdot 4 = \frac{1}{4} \cdot 4

化简

\frac{3 π}{4} θ \cdot 4 + 2 πn θ \cdot 4 = \frac{1}{4} \cdot 4

化简

\frac{3 π}{4} θ \cdot 4 : 3 π θ

\frac{3 π}{4} θ \cdot 4

分式相乘:

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

= \frac{3 \cdot 4 π}{4} θ

约分:

4

= θ \cdot 3 π

化简

2 πn θ \cdot 4 : 8 πn θ

2 πn θ \cdot 4

数字相乘:

2 \cdot 4 = 8

= 8 πn θ

化简

\frac{1}{4} \cdot 4 : 1

\frac{1}{4} \cdot 4

将项转换为分式:

4 = \frac{4}{1}

= \frac{1}{4} \cdot \frac{4}{1}

交叉约分:

4

= 1

3 π θ + 8 πn θ = 1

3 π θ + 8 πn θ = 1

3 π θ + 8 πn θ = 1

分解

3 π θ + 8 πn θ : π θ (3 + 8 n)

3 π θ + 8 πn θ

因式分解出通项

θ π

= θ π (3 + 8 n)

π θ (3 + 8 n) = 1

两边除以

π (3 + 8 n); n \neq = - \frac{3}{8}

π θ (3 + 8 n) = 1

两边除以

π (3 + 8 n); n \neq = - \frac{3}{8}

\frac{π θ ( 3 + 8 n )}{π ( 3 + 8 n )} = \frac{1}{π ( 3 + 8 n )}; n \neq = - \frac{3}{8}

化简

θ = \frac{1}{π ( 3 + 8 n )}; n \neq = - \frac{3}{8}

θ = \frac{1}{π ( 3 + 8 n )}; n \neq = - \frac{3}{8}

解

\frac{1}{4 θ} = \frac{5 π}{4} + 2 πn : θ = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{5}{8}

\frac{1}{4 θ} = \frac{5 π}{4} + 2 πn

在两边乘以

θ

\frac{1}{4 θ} = \frac{5 π}{4} + 2 πn

在两边乘以

θ

\frac{1}{4 θ} θ = \frac{5 π}{4} θ + 2 πn θ

化简

\frac{1}{4} = \frac{5 π}{4} θ + 2 πn θ

\frac{1}{4} = \frac{5 π}{4} θ + 2 πn θ

交换两边

\frac{5 π}{4} θ + 2 πn θ = \frac{1}{4}

在两边乘以

4

\frac{5 π}{4} θ + 2 πn θ = \frac{1}{4}

在两边乘以

4

\frac{5 π}{4} θ \cdot 4 + 2 πn θ \cdot 4 = \frac{1}{4} \cdot 4

化简

\frac{5 π}{4} θ \cdot 4 + 2 πn θ \cdot 4 = \frac{1}{4} \cdot 4

化简

\frac{5 π}{4} θ \cdot 4 : 5 π θ

\frac{5 π}{4} θ \cdot 4

分式相乘:

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

= \frac{5 \cdot 4 π}{4} θ

约分:

4

= θ \cdot 5 π

化简

2 πn θ \cdot 4 : 8 πn θ

2 πn θ \cdot 4

数字相乘:

2 \cdot 4 = 8

= 8 πn θ

化简

\frac{1}{4} \cdot 4 : 1

\frac{1}{4} \cdot 4

将项转换为分式:

4 = \frac{4}{1}

= \frac{1}{4} \cdot \frac{4}{1}

交叉约分:

4

= 1

5 π θ + 8 πn θ = 1

5 π θ + 8 πn θ = 1

5 π θ + 8 πn θ = 1

分解

5 π θ + 8 πn θ : π θ (5 + 8 n)

5 π θ + 8 πn θ

因式分解出通项

θ π

= θ π (5 + 8 n)

π θ (5 + 8 n) = 1

两边除以

π (5 + 8 n); n \neq = - \frac{5}{8}

π θ (5 + 8 n) = 1

两边除以

π (5 + 8 n); n \neq = - \frac{5}{8}

\frac{π θ ( 5 + 8 n )}{π ( 5 + 8 n )} = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{5}{8}

化简

θ = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{5}{8}

θ = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{5}{8}

θ = \frac{1}{π ( 3 + 8 n )}, θ = \frac{1}{π ( 5 + 8 n )}; n \neq = - \frac{3}{8}, n \neq = - \frac{5}{8}

作图

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