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受欢迎的三角函数 >

-cos(20.02t)=sin(20t)

解答

- cos (20.02 t) = sin (20 t)

解答

t = 25 π (4 n + 1), t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

+1

度数

t = 450 0^{\circ} + 1800 0^{\circ} n, t = - 2.24887 \dots^{\circ} - 8.99550 \dots^{\circ} n

求解步骤

- cos (20.02 t) = sin (20 t)

乘以

- 1

cos (20.02 t) = - sin (20 t)

使用三角恒等式改写

cos (20.02 t) = - sin (20 t)

利用以下特性:

- sin (x) = sin (- x)

cos (20.02 t) = sin (- (20 t))

利用以下特性:

cos (x) = sin (\frac{π}{2} - x)

sin (\frac{π}{2} - 20.02 t) = sin (- (20 t))

sin (\frac{π}{2} - 20.02 t) = sin (- (20 t))

使用反三角函数性质

sin (\frac{π}{2} - 20.02 t) = sin (- (20 t))

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

- (20 t) = \frac{π}{2} - 20.02 t + 2 πn, - (20 t) = π - (\frac{π}{2} - 20.02 t) + 2 πn

- (20 t) = \frac{π}{2} - 20.02 t + 2 πn, - (20 t) = π - (\frac{π}{2} - 20.02 t) + 2 πn

- (20 t) = \frac{π}{2} - 20.02 t + 2 πn : t = 25 π (4 n + 1)

- (20 t) = \frac{π}{2} - 20.02 t + 2 πn

在两边乘以

100

- (20 t) = \frac{π}{2} - 20.02 t + 2 πn

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere are

2

digits to the right of the decimal point, therefore multiply by

100

- 20 t \cdot 100 = \frac{π}{2} \cdot 100 - 20.02 t \cdot 100 + 2 πn \cdot 100

整理后得

- 2000 t = 50 π - 2002 t + 200 πn

- 2000 t = 50 π - 2002 t + 200 πn

将

2002 t

para o lado esquerdo

- 2000 t = 50 π - 2002 t + 200 πn

两边加上

2002 t

- 2000 t + 2002 t = 50 π - 2002 t + 200 πn + 2002 t

化简

2 t = 50 π + 200 πn

2 t = 50 π + 200 πn

两边除以

2

2 t = 50 π + 200 πn

两边除以

2

\frac{2 t}{2} = \frac{50 π}{2} + \frac{200 πn}{2}

化简

\frac{2 t}{2} = \frac{50 π}{2} + \frac{200 πn}{2}

化简

\frac{2 t}{2} : t

\frac{2 t}{2}

数字相除:

\frac{2}{2} = 1

= t

化简

\frac{50 π}{2} + \frac{200 πn}{2} : 25 π (4 n + 1)

\frac{50 π}{2} + \frac{200 πn}{2}

使用法则

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

= \frac{50 π + 200 πn}{2}

分解

50 π + 200 πn : 50 π (1 + 4 n)

50 π + 200 πn

改写为

= 1 \cdot 50 π + 4 \cdot 50 πn

因式分解出通项

50 π

= 50 π (1 + 4 n)

= \frac{50 π ( 1 + 4 n )}{2}

数字相除:

\frac{50}{2} = 25

= 25 π (4 n + 1)

t = 25 π (4 n + 1)

t = 25 π (4 n + 1)

t = 25 π (4 n + 1)

- (20 t) = π - (\frac{π}{2} - 20.02 t) + 2 πn : t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

- (20 t) = π - (\frac{π}{2} - 20.02 t) + 2 πn

展开

- (20 t) : - 20 t

- (20 t)

去除括号:

(a) = a

= - 20 t

展开

π - (\frac{π}{2} - 20.02 t) + 2 πn : π - \frac{π}{2} + 20.02 t + 6.28318 \dots n

π - (\frac{π}{2} - 20.02 t) + 2 πn

数字相乘:

2 \cdot 3.14159 \dots = 6.28318 \dots

= π - (- 20.02 t + \frac{π}{2}) + 6.28318 \dots n

- (\frac{π}{2} - 20.02 t) : - \frac{π}{2} + 20.02 t

- (\frac{π}{2} - 20.02 t)

打开括号

= - (\frac{π}{2}) - (- 20.02 t)

使用加减运算法则

- (- a) = a, - (a) = - a

= - \frac{π}{2} + 20.02 t

= π - \frac{π}{2} + 20.02 t + 6.28318 \dots n

- 20 t = π - \frac{π}{2} + 20.02 t + 6.28318 \dots n

将

20.02 t

para o lado esquerdo

- 20 t = π - \frac{π}{2} + 20.02 t + 6.28318 \dots n

两边减去

20.02 t

- 20 t - 20.02 t = π - \frac{π}{2} + 20.02 t + 6.28318 \dots n - 20.02 t

化简

- 40.02 t = π - \frac{π}{2} + 6.28318 \dots n

- 40.02 t = π - \frac{π}{2} + 6.28318 \dots n

两边除以

- 40.02

- 40.02 t = π - \frac{π}{2} + 6.28318 \dots n

两边除以

- 40.02

\frac{- 40.02 t}{- 40.02} = \frac{π}{- 40.02} - \frac{\frac{π}{2}}{- 40.02} + \frac{6.28318 \dots n}{- 40.02}

化简

\frac{- 40.02 t}{- 40.02} = \frac{π}{- 40.02} - \frac{\frac{π}{2}}{- 40.02} + \frac{6.28318 \dots n}{- 40.02}

化简

\frac{- 40.02 t}{- 40.02} : t

\frac{- 40.02 t}{- 40.02}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{40.02 t}{40.02}

约分:

40.02

= t

化简

\frac{π}{- 40.02} - \frac{\frac{π}{2}}{- 40.02} + \frac{6.28318 \dots n}{- 40.02} : - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

\frac{π}{- 40.02} - \frac{\frac{π}{2}}{- 40.02} + \frac{6.28318 \dots n}{- 40.02}

使用法则

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

= \frac{π - \frac{π}{2} + 6.28318 \dots n}{- 40.02}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{π - \frac{π}{2} + 6.28318 \dots n}{40.02}

化简

π - \frac{π}{2} + 6.28318 \dots n : 6.28318 \dots n + \frac{π}{2}

π - \frac{π}{2} + 6.28318 \dots n

将项转换为分式:

π = \frac{π 2}{2}

= \frac{π 2}{2} - \frac{π}{2}

因为分母相等，所以合并分式:

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

= \frac{π 2 - π}{2}

同类项相加:

2 π - π = π

= \frac{π}{2}

= - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

t = 25 π (4 n + 1), t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

t = 25 π (4 n + 1), t = - \frac{6.28318 \dots n + \frac{π}{2}}{40.02}

作图

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