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受欢迎的三角函数 >

0=1.9cos(0.1t)-1.25sin(0.2t)

解答

0 = 1.9 cos (0.1 t) - 1.25 sin (0.2 t)

解答

t = 5 π + 20 πn, t = 15 π + 20 πn, t = \frac{0.86331 \dots + 2 πn}{0.1}, t = \frac{π - 0.86331 \dots + 2 πn}{0.1}

度数

t = 90 0^{\circ} + 360 0^{\circ} n, t = 270 0^{\circ} + 360 0^{\circ} n, t = 494.64197 \dots^{\circ} + 360 0^{\circ} n, t = 1305.35802 \dots^{\circ} + 360 0^{\circ} n

求解步骤

0 = 1.9 cos (0.1 t) - 1.25 sin (0.2 t)

交换两边

1.9 cos (0.1 t) - 1.25 sin (0.2 t) = 0

令

: u = 0.1 t

1.9 cos (u) - 1.25 sin (2 u) = 0

使用三角恒等式改写

- 1.25 sin (2 u) + 1.9 cos (u)

使用倍角公式:

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

= - 1.25 \cdot 2 sin (u) cos (u) + 1.9 cos (u)

化简

= - 2.5 sin (u) cos (u) + 1.9 cos (u)

1.9 cos (u) - 2.5 cos (u) sin (u) = 0

分解

1.9 cos (u) - 2.5 cos (u) sin (u) : - cos (u) (2.5 sin (u) - 1.9)

1.9 cos (u) - 2.5 cos (u) sin (u)

因式分解出通项

- cos (u)

= - cos (u) (- 1.9 + 2.5 sin (u))

- cos (u) (2.5 sin (u) - 1.9) = 0

分别求解每个部分

cos (u) = 0 or 2.5 sin (u) - 1.9 = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

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

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

2.5 sin (u) - 1.9 = 0 : u = arcsin (\frac{19}{25}) + 2 πn, u = π - arcsin (\frac{19}{25}) + 2 πn

2.5 sin (u) - 1.9 = 0

在两边乘以

10

2.5 sin (u) - 1.9 = 0

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

2.5 sin (u) \cdot 10 - 1.9 \cdot 10 = 0 \cdot 10

整理后得

25 sin (u) - 19 = 0

25 sin (u) - 19 = 0

将

19

到右边

25 sin (u) - 19 = 0

两边加上

19

25 sin (u) - 19 + 19 = 0 + 19

化简

25 sin (u) = 19

25 sin (u) = 19

两边除以

25

25 sin (u) = 19

两边除以

25

\frac{25 sin ( u )}{25} = \frac{19}{25}

化简

sin (u) = \frac{19}{25}

sin (u) = \frac{19}{25}

使用反三角函数性质

sin (u) = \frac{19}{25}

sin (u) = \frac{19}{25}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

u = arcsin (\frac{19}{25}) + 2 πn, u = π - arcsin (\frac{19}{25}) + 2 πn

u = arcsin (\frac{19}{25}) + 2 πn, u = π - arcsin (\frac{19}{25}) + 2 πn

合并所有解

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = arcsin (\frac{19}{25}) + 2 πn, u = π - arcsin (\frac{19}{25}) + 2 πn

u = 0.1 t

代回

0.1 t = \frac{π}{2} + 2 πn : t = 5 π + 20 πn

0.1 t = \frac{π}{2} + 2 πn

在两边乘以

10

0.1 t = \frac{π}{2} + 2 πn

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.1 t \cdot 10 = \frac{π}{2} \cdot 10 + 2 πn \cdot 10

整理后得

t = 5 π + 20 πn

t = 5 π + 20 πn

0.1 t = \frac{3 π}{2} + 2 πn : t = 15 π + 20 πn

0.1 t = \frac{3 π}{2} + 2 πn

在两边乘以

10

0.1 t = \frac{3 π}{2} + 2 πn

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.1 t \cdot 10 = \frac{3 π}{2} \cdot 10 + 2 πn \cdot 10

整理后得

t = 15 π + 20 πn

t = 15 π + 20 πn

0.1 t = arcsin (\frac{19}{25}) + 2 πn : t = \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

0.1 t = arcsin (\frac{19}{25}) + 2 πn

两边除以

0.1

0.1 t = arcsin (\frac{19}{25}) + 2 πn

两边除以

0.1

\frac{0.1 t}{0.1} = \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

化简

\frac{0.1 t}{0.1} = \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

化简

\frac{0.1 t}{0.1} : t

\frac{0.1 t}{0.1}

约分:

0.1

= t

化简

\frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1} : \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

\frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

使用法则

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

= \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

0.1 t = π - arcsin (\frac{19}{25}) + 2 πn : t = \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

0.1 t = π - arcsin (\frac{19}{25}) + 2 πn

两边除以

0.1

0.1 t = π - arcsin (\frac{19}{25}) + 2 πn

两边除以

0.1

\frac{0.1 t}{0.1} = \frac{π}{0.1} - \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

化简

\frac{0.1 t}{0.1} = \frac{π}{0.1} - \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

化简

\frac{0.1 t}{0.1} : t

\frac{0.1 t}{0.1}

约分:

0.1

= t

化简

\frac{π}{0.1} - \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1} : \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

\frac{π}{0.1} - \frac{arcsin ( \frac{19}{25} )}{0.1} + \frac{2 πn}{0.1}

使用法则

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

= \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

t = 5 π + 20 πn, t = 15 π + 20 πn, t = \frac{arcsin ( \frac{19}{25} ) + 2 πn}{0.1}, t = \frac{π - arcsin ( \frac{19}{25} ) + 2 πn}{0.1}

以小数形式表示解

t = 5 π + 20 πn, t = 15 π + 20 πn, t = \frac{0.86331 \dots + 2 πn}{0.1}, t = \frac{π - 0.86331 \dots + 2 πn}{0.1}

作图

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