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受欢迎的三角函数 >

sin^3(t)=sin(t)

解答

sin^{3} (t) = sin (t)

解答

t = 2 πn, t = π + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

+1

度数

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{3} (t) = sin (t)

用替代法求解

sin^{3} (t) = sin (t)

令

: sin (t) = u

u^{3} = u

u^{3} = u : u = 0, u = - 1, u = 1

u^{3} = u

将

u

para o lado esquerdo

u^{3} = u

两边减去

u

u^{3} - u = u - u

化简

u^{3} - u = 0

u^{3} - u = 0

因式分解

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

因式分解出通项

u : u (u^{2} - 1)

u^{3} - u

使用指数法则:

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

u^{3} = u^{2} u

= u^{2} u - u

因式分解出通项

u

= u (u^{2} - 1)

= u (u^{2} - 1)

分解

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

将

1

改写为

1^{2}

= u^{2} - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解为

u = 0, u = - 1, u = 1

u = sin (t)

代回

sin (t) = 0, sin (t) = - 1, sin (t) = 1

sin (t) = 0, sin (t) = - 1, sin (t) = 1

sin (t) = 0 : t = 2 πn, t = π + 2 πn

sin (t) = 0

sin (t) = 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}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

解

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

sin (t) = - 1

sin (t) = - 1

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}

t = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

sin (t) = 1 : t = \frac{π}{2} + 2 πn

sin (t) = 1

sin (t) = 1

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}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

合并所有解

t = 2 πn, t = π + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{2} + 2 πn

作图

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