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受欢迎的三角函数 >

3sin(t)=-3+cos(t)

解答

3 sin (t) = - 3 + cos (t)

解答

t = \frac{3 π}{2} + 2 πn, t = 2 π - 0.92729 \dots + 2 πn

+1

度数

t = 27 0^{\circ} + 36 0^{\circ} n, t = 306.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (t) = - 3 + cos (t)

两边进行平方

(3 sin (t))^{2} = (- 3 + cos (t))^{2}

两边减去

(- 3 + cos (t))^{2}

9 sin^{2} (t) - 9 + 6 cos (t) - cos^{2} (t) = 0

使用三角恒等式改写

- 9 - cos^{2} (t) + 6 cos (t) + 9 sin^{2} (t)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= - 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t))

化简

- 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t)) : 6 cos (t) - 10 cos^{2} (t)

- 9 - cos^{2} (t) + 6 cos (t) + 9 (1 - cos^{2} (t))

乘开

9 (1 - cos^{2} (t)) : 9 - 9 cos^{2} (t)

9 (1 - cos^{2} (t))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (t)

= 9 \cdot 1 - 9 cos^{2} (t)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (t)

= - 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t)

化简

- 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t) : 6 cos (t) - 10 cos^{2} (t)

- 9 - cos^{2} (t) + 6 cos (t) + 9 - 9 cos^{2} (t)

对同类项分组

= - cos^{2} (t) + 6 cos (t) - 9 cos^{2} (t) - 9 + 9

同类项相加:

- cos^{2} (t) - 9 cos^{2} (t) = - 10 cos^{2} (t)

= - 10 cos^{2} (t) + 6 cos (t) - 9 + 9

- 9 + 9 = 0

= 6 cos (t) - 10 cos^{2} (t)

= 6 cos (t) - 10 cos^{2} (t)

= 6 cos (t) - 10 cos^{2} (t)

- 10 cos^{2} (t) + 6 cos (t) = 0

用替代法求解

- 10 cos^{2} (t) + 6 cos (t) = 0

令

: cos (t) = u

- 10 u^{2} + 6 u = 0

- 10 u^{2} + 6 u = 0 : u = 0, u = \frac{3}{5}

- 10 u^{2} + 6 u = 0

使用求根公式求解

- 10 u^{2} + 6 u = 0

二次方程求根公式:

若

a = - 10, b = 6, c = 0

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

6^{2} - 4 (- 10) \cdot 0 = 6

6^{2} - 4 (- 10) \cdot 0

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 10 \cdot 0

使用法则

0 \cdot a = 0

= 6^{2} + 0

6^{2} + 0 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 6

u_{1, 2} = \frac{- 6 \pm 6}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- 6 + 6}{2 ( - 10 )}, u_{2} = \frac{- 6 - 6}{2 ( - 10 )}

u = \frac{- 6 + 6}{2 ( - 10 )} : 0

\frac{- 6 + 6}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 6 + 6}{- 2 \cdot 10}

数字相加/相减:

- 6 + 6 = 0

= \frac{0}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{0}{- 20}

使用分式法则:

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

= - \frac{0}{20}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 6 - 6}{2 ( - 10 )} : \frac{3}{5}

\frac{- 6 - 6}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 6 - 6}{- 2 \cdot 10}

数字相减:

- 6 - 6 = - 12

= \frac{- 12}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 12}{- 20}

使用分式法则:

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

= \frac{12}{20}

约分:

4

= \frac{3}{5}

二次方程组的解是:

u = 0, u = \frac{3}{5}

u = cos (t)

代回

cos (t) = 0, cos (t) = \frac{3}{5}

cos (t) = 0, cos (t) = \frac{3}{5}

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

cos (t) = 0

cos (t) = 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}

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

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

cos (t) = \frac{3}{5} : t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

cos (t) = \frac{3}{5}

使用反三角函数性质

cos (t) = \frac{3}{5}

cos (t) = \frac{3}{5}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

合并所有解

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = arccos (\frac{3}{5}) + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

将解代入原方程进行验证

将它们代入

3 sin (t) = - 3 + cos (t)

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

假

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

3 sin (t) = - 3 + cos (t)

代入

t = \frac{π}{2} + 2 π 1

3 sin (\frac{π}{2} + 2 π 1) = - 3 + cos (\frac{π}{2} + 2 π 1)

整理后得

3 = - 3

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

真

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

3 sin (t) = - 3 + cos (t)

代入

t = \frac{3 π}{2} + 2 π 1

3 sin (\frac{3 π}{2} + 2 π 1) = - 3 + cos (\frac{3 π}{2} + 2 π 1)

整理后得

- 3 = - 3

\Rightarrow 真

检验

arccos (\frac{3}{5}) + 2 πn

的解:

假

arccos (\frac{3}{5}) + 2 πn

代入

n = 1

arccos (\frac{3}{5}) + 2 π 1

对于

3 sin (t) = - 3 + cos (t)

代入

t = arccos (\frac{3}{5}) + 2 π 1

3 sin (arccos (\frac{3}{5}) + 2 π 1) = - 3 + cos (arccos (\frac{3}{5}) + 2 π 1)

整理后得

2.4 = - 2.4

\Rightarrow 假

检验

2 π - arccos (\frac{3}{5}) + 2 πn

的解:

真

2 π - arccos (\frac{3}{5}) + 2 πn

代入

n = 1

2 π - arccos (\frac{3}{5}) + 2 π 1

对于

3 sin (t) = - 3 + cos (t)

代入

t = 2 π - arccos (\frac{3}{5}) + 2 π 1

3 sin (2 π - arccos (\frac{3}{5}) + 2 π 1) = - 3 + cos (2 π - arccos (\frac{3}{5}) + 2 π 1)

整理后得

- 2.4 = - 2.4

\Rightarrow 真

t = \frac{3 π}{2} + 2 πn, t = 2 π - arccos (\frac{3}{5}) + 2 πn

以小数形式表示解

t = \frac{3 π}{2} + 2 πn, t = 2 π - 0.92729 \dots + 2 πn

作图

流行的例子

(tan(x))/3 =sqrt(3)

\frac{tan ( x )}{3} = 3

cos(x)=0.45,cos(pi-x)

cos (x) = 0.45, cos (π - x)

4cos^2(x)+9sin(x)-9=0

4 cos^{2} (x) + 9 sin (x) - 9 = 0

tan(θ)=-12/5 , pi/2 <= θ<= pi

tan (θ) = - \frac{12}{5}, \frac{π}{2} \leq θ \leq π

cos(x)=sin(2x+pi/2),0<= x<= 2pi

cos (x) = sin (2 x + \frac{π}{2}), 0 \leq x \leq 2 π