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受欢迎的三角函数 >

sin(θ)=1-cos(θ)

解答

sin (θ) = 1 - cos (θ)

解答

θ = 2 πn, θ = 2 πn + \frac{π}{2}

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (θ) = 1 - cos (θ)

两边减去

1 - cos (θ)

sin (θ) - 1 + cos (θ) = 0

使用三角恒等式改写

sin (θ) - 1 + cos (θ)

sin (θ) + cos (θ) = 2 sin (θ + \frac{π}{4})

sin (θ) + cos (θ)

改写为

= 2 (\frac{1}{2} sin (θ) + \frac{1}{2} cos (θ))

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (θ) + sin (\frac{π}{4}) cos (θ))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (θ + \frac{π}{4})

= - 1 + 2 sin (θ + \frac{π}{4})

- 1 + 2 sin (θ + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (θ + \frac{π}{4}) = 0

两边加上

1

- 1 + 2 sin (θ + \frac{π}{4}) + 1 = 0 + 1

化简

2 sin (θ + \frac{π}{4}) = 1

2 sin (θ + \frac{π}{4}) = 1

两边除以

2

2 sin (θ + \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( θ + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( θ + \frac{π}{4} )}{2} : sin (θ + \frac{π}{4})

\frac{2 sin ( θ + \frac{π}{4} )}{2}

约分:

2

= sin (θ + \frac{π}{4})

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 (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \frac{2}{2}

sin (θ + \frac{π}{4}) = \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}

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

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

解

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

θ + \frac{π}{4} = \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

θ + \frac{π}{4} - \frac{π}{4} = \frac{π}{4} + 2 πn - \frac{π}{4}

化简

θ = 2 πn

解

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

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

将

\frac{π}{4}

到右边

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

两边减去

\frac{π}{4}

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

化简

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

化简

θ + \frac{π}{4} - \frac{π}{4} : θ

θ + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= θ

化简

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

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

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

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn + \frac{π}{2}

θ = 2 πn, θ = 2 πn + \frac{π}{2}

作图

流行的例子

cos(θ)-2sin(θ)cos(θ)=0

cos (θ) - 2 sin (θ) cos (θ) = 0

solvefor x,cos(xy)=sin(x+y)

so l v e f or x, cos (x y) = sin (x + y)

cot (x) = \frac{3}{4}

3sin(x)-cos^2(x)+3=0

3 sin (x) - cos^{2} (x) + 3 = 0

tan^2(x)-3=0,0<x< pi/2

tan^{2} (x) - 3 = 0, 0 < x < \frac{π}{2}