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受欢迎的三角函数 >

-(2)cos(10t)=(6)sin(10t)

解答

- (2) cos (10 t) = (6) sin (10 t)

解答

t = - \frac{0.32175 \dots}{10} + \frac{πn}{10}

+1

度数

t = - 1.84349 \dots^{\circ} + 1 8^{\circ} n

求解步骤

- (2) cos (10 t) = (6) sin (10 t)

使用三角恒等式改写

- 2 cos (10 t) = 6 sin (10 t)

在两边除以

cos (10 t), cos (10 t) \neq = 0

\frac{- 2 cos ( 10 t )}{cos ( 10 t )} = \frac{6 sin ( 10 t )}{cos ( 10 t )}

化简

- 2 = \frac{6 sin ( 10 t )}{cos ( 10 t )}

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

- 2 = 6 tan (10 t)

交换两边

6 tan (10 t) = - 2

6 tan (10 t) = - 2

两边除以

6

6 tan (10 t) = - 2

两边除以

6

\frac{6 tan ( 10 t )}{6} = \frac{- 2}{6}

化简

\frac{6 tan ( 10 t )}{6} = \frac{- 2}{6}

化简

\frac{6 tan ( 10 t )}{6} : tan (10 t)

\frac{6 tan ( 10 t )}{6}

数字相除:

\frac{6}{6} = 1

= tan (10 t)

化简

\frac{- 2}{6} : - \frac{1}{3}

\frac{- 2}{6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

tan (10 t) = - \frac{1}{3}

tan (10 t) = - \frac{1}{3}

tan (10 t) = - \frac{1}{3}

使用反三角函数性质

tan (10 t) = - \frac{1}{3}

tan (10 t) = - \frac{1}{3}

的通解

tan (x) = - a \Rightarrow x = arctan (- a) + πn

10 t = arctan (- \frac{1}{3}) + πn

10 t = arctan (- \frac{1}{3}) + πn

解

10 t = arctan (- \frac{1}{3}) + πn : t = - \frac{arctan ( \frac{1}{3} )}{10} + \frac{πn}{10}

10 t = arctan (- \frac{1}{3}) + πn

化简

arctan (- \frac{1}{3}) + πn : - arctan (\frac{1}{3}) + πn

arctan (- \frac{1}{3}) + πn

利用以下特性:

arctan (- x) = - arctan (x)

arctan (- \frac{1}{3}) = - arctan (\frac{1}{3})

= - arctan (\frac{1}{3}) + πn

10 t = - arctan (\frac{1}{3}) + πn

两边除以

10

10 t = - arctan (\frac{1}{3}) + πn

两边除以

10

\frac{10 t}{10} = - \frac{arctan ( \frac{1}{3} )}{10} + \frac{πn}{10}

化简

t = - \frac{arctan ( \frac{1}{3} )}{10} + \frac{πn}{10}

t = - \frac{arctan ( \frac{1}{3} )}{10} + \frac{πn}{10}

t = - \frac{arctan ( \frac{1}{3} )}{10} + \frac{πn}{10}

以小数形式表示解

t = - \frac{0.32175 \dots}{10} + \frac{πn}{10}

作图

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