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受欢迎的三角函数 >

sin(θ/2)=-1/(sqrt(2))

解答

sin (\frac{θ}{2}) = - \frac{1}{2}

解答

θ = \frac{5 π}{2} + 4 πn, θ = \frac{7 π}{2} + 4 πn

求解步骤

sin (\frac{θ}{2}) = - \frac{1}{2}

化简

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

sin (\frac{θ}{2}) = - \frac{2}{2}

sin (\frac{θ}{2}) = - \frac{2}{2}

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}

\frac{θ}{2} = \frac{5 π}{4} + 2 πn, \frac{θ}{2} = \frac{7 π}{4} + 2 πn

\frac{θ}{2} = \frac{5 π}{4} + 2 πn, \frac{θ}{2} = \frac{7 π}{4} + 2 πn

解

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

\frac{θ}{2} = \frac{5 π}{4} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{5 π}{4} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn : \frac{5 π}{2} + 4 πn

2 \cdot \frac{5 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{4} = \frac{5 π}{2}

2 \cdot \frac{5 π}{4}

分式相乘:

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

= \frac{5 π 2}{4}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{4}

约分:

2

= \frac{5 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn

解

\frac{θ}{2} = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{2} + 4 πn

\frac{θ}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

\frac{θ}{2} = \frac{7 π}{4} + 2 πn

在两边乘以

2

\frac{2 θ}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} = 2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn : \frac{7 π}{2} + 4 πn

2 \cdot \frac{7 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{4} = \frac{7 π}{2}

2 \cdot \frac{7 π}{4}

分式相乘:

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

= \frac{7 π 2}{4}

数字相乘:

7 \cdot 2 = 14

= \frac{14 π}{4}

约分:

2

= \frac{7 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{7 π}{2} + 4 πn

θ = \frac{5 π}{2} + 4 πn, θ = \frac{7 π}{2} + 4 πn

作图

流行的例子

sec^2(x)=sqrt(3)tan(x)+1

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

-3cos(2θ)-sin(θ)-6=-8

- 3 cos (2 θ) - sin (θ) - 6 = - 8

sin^2(x)+1= 7/2 cos^2(x)

sin^{2} (x) + 1 = \frac{7}{2} cos^{2} (x)

cos(θ)sin(θ)-2cos(θ)=0

cos (θ) sin (θ) - 2 cos (θ) = 0

solvefor y,x=tan^2(y)

so l v e f or y, x = tan^{2} (y)