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受欢迎的三角函数 >

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

解答

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

解答

x = 4 πn, x = 2 π + 4 πn

+1

度数

x = 0^{\circ} + 72 0^{\circ} n, x = 36 0^{\circ} + 72 0^{\circ} n

求解步骤

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

两边减去

1 - cos (x)

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

令

: u = \frac{x}{2}

sin (2 u) sin (u) - 1 + cos (2 u) = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

= sin (2 u) sin (u) - 2 sin^{2} (u)

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

分解

- 2 sin^{2} (u) + sin (2 u) sin (u) : sin (u) (- 2 sin (u) + sin (2 u))

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

使用指数法则:

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

sin^{2} (u) = sin (u) sin (u)

= - 2 sin (u) sin (u) + sin (u) sin (2 u)

因式分解出通项

sin (u)

= sin (u) (- 2 sin (u) + sin (2 u))

sin (u) (- 2 sin (u) + sin (2 u)) = 0

分别求解每个部分

sin (u) = 0 or - 2 sin (u) + sin (2 u) = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

- 2 sin (u) + sin (2 u) = 0 : u = 2 πn, u = π + 2 πn

- 2 sin (u) + sin (2 u) = 0

使用三角恒等式改写

sin (2 u) - 2 sin (u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (u) cos (u) - 2 sin (u)

- 2 sin (u) + 2 cos (u) sin (u) = 0

分解

- 2 sin (u) + 2 cos (u) sin (u) : 2 sin (u) (cos (u) - 1)

- 2 sin (u) + 2 cos (u) sin (u)

因式分解出通项

2 sin (u)

= 2 sin (u) (- 1 + cos (u))

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

分别求解每个部分

sin (u) = 0 or cos (u) - 1 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

cos (u) - 1 = 0 : u = 2 πn

cos (u) - 1 = 0

将

1

到右边

cos (u) - 1 = 0

两边加上

1

cos (u) - 1 + 1 = 0 + 1

化简

cos (u) = 1

cos (u) = 1

cos (u) = 1

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}

u = 0 + 2 πn

u = 0 + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn

合并所有解

u = 2 πn, u = π + 2 πn

合并所有解

u = 2 πn, u = π + 2 πn

u = \frac{x}{2}

代回

\frac{x}{2} = 2 πn : x = 4 πn

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 πn

化简

x = 4 πn

x = 4 πn

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

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{x}{2} = π + 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 π + 2 \cdot 2 πn

化简

x = 2 π + 4 πn

x = 2 π + 4 πn

x = 4 πn, x = 2 π + 4 πn

作图

流行的例子

2 sin (x - \frac{π}{3}) = 1

tan (θ) = 2.7

4 csc (θ) + 4 = 0

tan (a) = \frac{7}{8}

(sec(x)sin(x))/(tan(x)cot(x))=sin^2(x)

\frac{sec ( x ) sin ( x )}{tan ( x ) cot ( x )} = sin^{2} (x)