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受欢迎的三角函数 >

sin(0.5pi)cos(0.5pi)=0.5sin(2x+1)

解答

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

解答

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

+1

度数

x = - 28.64788 \dots^{\circ} + 18 0^{\circ} n, x = 61.35211 \dots^{\circ} + 18 0^{\circ} n

求解步骤

sin (0.5 π) cos (0.5 π) = 0.5 sin (2 x + 1)

sin (0.5 π) = 1

sin (0.5 π)

= sin (\frac{1}{2} π)

化简:

\frac{1}{2} π = \frac{π}{2}

\frac{1}{2} π

分式相乘:

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

= \frac{1 π}{2}

乘以:

1 π = π

= \frac{π}{2}

= sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

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}

= 1

= 1

cos (0.5 π) = 0

cos (0.5 π)

= cos (\frac{1}{2} π)

化简:

\frac{1}{2} π = \frac{π}{2}

\frac{1}{2} π

分式相乘:

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

= \frac{1 π}{2}

乘以:

1 π = π

= \frac{π}{2}

= cos (\frac{π}{2})

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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}

= 0

= 0

1 \cdot 0 = 0.5 sin (2 x + 1)

交换两边

0.5 sin (2 x + 1) = 1 \cdot 0

使用法则

0 \cdot a = 0

0.5 sin (2 x + 1) = 0

在两边乘以

10

0.5 sin (2 x + 1) = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.5 sin (2 x + 1) \cdot 10 = 0 \cdot 10

整理后得

5 sin (2 x + 1) = 0

5 sin (2 x + 1) = 0

两边除以

5

5 sin (2 x + 1) = 0

两边除以

5

\frac{5 sin ( 2 x + 1 )}{5} = \frac{0}{5}

化简

sin (2 x + 1) = 0

sin (2 x + 1) = 0

sin (2 x + 1) = 0

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}

2 x + 1 = 0 + 2 πn, 2 x + 1 = π + 2 πn

2 x + 1 = 0 + 2 πn, 2 x + 1 = π + 2 πn

解

2 x + 1 = 0 + 2 πn : x = πn - \frac{1}{2}

2 x + 1 = 0 + 2 πn

0 + 2 πn = 2 πn

2 x + 1 = 2 πn

将

1

到右边

2 x + 1 = 2 πn

两边减去

1

2 x + 1 - 1 = 2 πn - 1

化简

2 x = 2 πn - 1

2 x = 2 πn - 1

两边除以

2

2 x = 2 πn - 1

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2} - \frac{1}{2}

化简

x = πn - \frac{1}{2}

x = πn - \frac{1}{2}

解

2 x + 1 = π + 2 πn : x = \frac{π}{2} - \frac{1}{2} + πn

2 x + 1 = π + 2 πn

将

1

到右边

2 x + 1 = π + 2 πn

两边减去

1

2 x + 1 - 1 = π + 2 πn - 1

化简

2 x = π + 2 πn - 1

2 x = π + 2 πn - 1

两边除以

2

2 x = π + 2 πn - 1

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

化简

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2} : \frac{π}{2} - \frac{1}{2} + πn

\frac{π}{2} + \frac{2 πn}{2} - \frac{1}{2}

对同类项分组

= \frac{π}{2} - \frac{1}{2} + \frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = \frac{π}{2} - \frac{1}{2} + πn

x = πn - \frac{1}{2}, x = \frac{π}{2} - \frac{1}{2} + πn

作图

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