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受欢迎的三角函数 >

solvefor t,v=311cos(30t+30)v

解答

求解

t, v = 311 cos (30 t + 3 0^{\circ}) v

解答

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{1.56758 \dots}{30}, t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{1.56758 \dots}{30}

+1

弧度

t = - \frac{π}{180} + \frac{1.56758 \dots}{30} + \frac{π}{15} n, t = \frac{π}{15} - \frac{π}{180} - \frac{1.56758 \dots}{30} + \frac{π}{15} n

求解步骤

v = 311 cos (30 t + 3 0^{\circ}) v

交换两边

311 cos (30 t + \frac{π}{6}) v = v

两边除以

311 v; v \neq = 0

311 cos (30 t + \frac{π}{6}) v = v

两边除以

311 v; v \neq = 0

\frac{311 cos ( 30 t + \frac{π}{6} ) v}{311 v} = \frac{v}{311 v}; v \neq = 0

化简

cos (30 t + \frac{π}{6}) = \frac{1}{311}; v \neq = 0

cos (30 t + \frac{π}{6}) = \frac{1}{311}; v \neq = 0

使用反三角函数性质

cos (30 t + 3 0^{\circ}) = \frac{1}{311}

cos (30 t + 3 0^{\circ}) = \frac{1}{311}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

30 t + 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n, 30 t + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n

30 t + 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n, 30 t + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n

解

30 t + 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n : t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

30 t + 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n

将

3 0^{\circ}

到右边

30 t + 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n

两边减去

3 0^{\circ}

30 t + 3 0^{\circ} - 3 0^{\circ} = arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

化简

30 t = arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

30 t = arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

两边除以

30

30 t = arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

两边除以

30

\frac{30 t}{30} = \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

化简

\frac{30 t}{30} = \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

化简

\frac{30 t}{30} : t

\frac{30 t}{30}

数字相除:

\frac{30}{30} = 1

= t

化简

\frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30} : \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

\frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

对同类项分组

= \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30} + \frac{arccos ( \frac{1}{311} )}{30}

\frac{36 0 ^{\circ} n}{30} = \frac{18 0 ^{\circ} n}{15}

\frac{36 0 ^{\circ} n}{30}

约分:

2

= \frac{18 0 ^{\circ} n}{15}

\frac{3 0 ^{\circ}}{30} = 1^{\circ}

\frac{3 0 ^{\circ}}{30}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{6 \cdot 30}

数字相乘:

6 \cdot 30 = 180

= 1^{\circ}

= \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}

解

30 t + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n : t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

30 t + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n

将

3 0^{\circ}

到右边

30 t + 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n

两边减去

3 0^{\circ}

30 t + 3 0^{\circ} - 3 0^{\circ} = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

化简

30 t = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

30 t = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

两边除以

30

30 t = 36 0^{\circ} - arccos (\frac{1}{311}) + 36 0^{\circ} n - 3 0^{\circ}

两边除以

30

\frac{30 t}{30} = 1 2^{\circ} - \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

化简

\frac{30 t}{30} = 1 2^{\circ} - \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

化简

\frac{30 t}{30} : t

\frac{30 t}{30}

数字相除:

\frac{30}{30} = 1

= t

化简

1 2^{\circ} - \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30} : 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

1 2^{\circ} - \frac{arccos ( \frac{1}{311} )}{30} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30}

对同类项分组

= 1 2^{\circ} + \frac{36 0 ^{\circ} n}{30} - \frac{3 0 ^{\circ}}{30} - \frac{arccos ( \frac{1}{311} )}{30}

1 2^{\circ} = 1 2^{\circ}

1 2^{\circ}

约分:

2

= 1 2^{\circ}

\frac{36 0 ^{\circ} n}{30} = \frac{18 0 ^{\circ} n}{15}

\frac{36 0 ^{\circ} n}{30}

约分:

2

= \frac{18 0 ^{\circ} n}{15}

\frac{3 0 ^{\circ}}{30} = 1^{\circ}

\frac{3 0 ^{\circ}}{30}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{6 \cdot 30}

数字相乘:

6 \cdot 30 = 180

= 1^{\circ}

= 1 2^{\circ} + \frac{18 0 ^{\circ} n}{15} - 1^{\circ} - \frac{arccos ( \frac{1}{311} )}{30}

对同类项分组

= 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{arccos ( \frac{1}{311} )}{30}, t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{arccos ( \frac{1}{311} )}{30}

以小数形式表示解

t = \frac{18 0 ^{\circ} n}{15} - 1^{\circ} + \frac{1.56758 \dots}{30}, t = 1 2^{\circ} - 1^{\circ} + \frac{18 0 ^{\circ} n}{15} - \frac{1.56758 \dots}{30}

作图

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