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受欢迎的三角函数 >

3sin(a)+cos(a)=1

解答

3 sin (a) + cos (a) = 1

解答

a = 2.49809 \dots + 2 πn, a = 2 πn

+1

度数

a = 143.13010 \dots^{\circ} + 36 0^{\circ} n, a = 0^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (a) + cos (a) = 1

两边减去

cos (a)

3 sin (a) = 1 - cos (a)

两边进行平方

(3 sin (a))^{2} = (1 - cos (a))^{2}

两边减去

(1 - cos (a))^{2}

9 sin^{2} (a) - 1 + 2 cos (a) - cos^{2} (a) = 0

使用三角恒等式改写

- 1 - cos^{2} (a) + 2 cos (a) + 9 sin^{2} (a)

使用毕达哥拉斯恒等式:

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

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

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

化简

- 1 - cos^{2} (a) + 2 cos (a) + 9 (1 - cos^{2} (a)) : 2 cos (a) - 10 cos^{2} (a) + 8

- 1 - cos^{2} (a) + 2 cos (a) + 9 (1 - cos^{2} (a))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

9 \cdot 1 = 9

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

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

化简

- 1 - cos^{2} (a) + 2 cos (a) + 9 - 9 cos^{2} (a) : 2 cos (a) - 10 cos^{2} (a) + 8

- 1 - cos^{2} (a) + 2 cos (a) + 9 - 9 cos^{2} (a)

对同类项分组

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

同类项相加:

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

= - 10 cos^{2} (a) + 2 cos (a) - 1 + 9

数字相加/相减:

- 1 + 9 = 8

= 2 cos (a) - 10 cos^{2} (a) + 8

= 2 cos (a) - 10 cos^{2} (a) + 8

= 2 cos (a) - 10 cos^{2} (a) + 8

8 - 10 cos^{2} (a) + 2 cos (a) = 0

用替代法求解

8 - 10 cos^{2} (a) + 2 cos (a) = 0

令

: cos (a) = u

8 - 10 u^{2} + 2 u = 0

8 - 10 u^{2} + 2 u = 0 : u = - \frac{4}{5}, u = 1

8 - 10 u^{2} + 2 u = 0

改写成标准形式

a x^{2} + b x + c = 0

- 10 u^{2} + 2 u + 8 = 0

使用求根公式求解

- 10 u^{2} + 2 u + 8 = 0

二次方程求根公式:

若

a = - 10, b = 2, c = 8

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 10 ) \cdot 8}{2 ( - 10 )}

2^{2} - 4 (- 10) \cdot 8 = 18

2^{2} - 4 (- 10) \cdot 8

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 10 \cdot 8

数字相乘:

4 \cdot 10 \cdot 8 = 320

= 2^{2} + 320

2^{2} = 4

= 4 + 320

数字相加:

4 + 320 = 324

= 324

因式分解数字:

324 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- 2 \pm 18}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- 2 + 18}{2 ( - 10 )}, u_{2} = \frac{- 2 - 18}{2 ( - 10 )}

u = \frac{- 2 + 18}{2 ( - 10 )} : - \frac{4}{5}

\frac{- 2 + 18}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 2 + 18}{- 2 \cdot 10}

数字相加/相减:

- 2 + 18 = 16

= \frac{16}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{16}{- 20}

使用分式法则:

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

= - \frac{16}{20}

约分:

4

= - \frac{4}{5}

u = \frac{- 2 - 18}{2 ( - 10 )} : 1

\frac{- 2 - 18}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 2 - 18}{- 2 \cdot 10}

数字相减:

- 2 - 18 = - 20

= \frac{- 20}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 20}{- 20}

使用分式法则:

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

= \frac{20}{20}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - \frac{4}{5}, u = 1

u = cos (a)

代回

cos (a) = - \frac{4}{5}, cos (a) = 1

cos (a) = - \frac{4}{5}, cos (a) = 1

cos (a) = - \frac{4}{5} : a = arccos (- \frac{4}{5}) + 2 πn, a = - arccos (- \frac{4}{5}) + 2 πn

cos (a) = - \frac{4}{5}

使用反三角函数性质

cos (a) = - \frac{4}{5}

cos (a) = - \frac{4}{5}

的通解

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

a = arccos (- \frac{4}{5}) + 2 πn, a = - arccos (- \frac{4}{5}) + 2 πn

a = arccos (- \frac{4}{5}) + 2 πn, a = - arccos (- \frac{4}{5}) + 2 πn

cos (a) = 1 : a = 2 πn

cos (a) = 1

cos (a) = 1

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}

a = 0 + 2 πn

a = 0 + 2 πn

解

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

合并所有解

a = arccos (- \frac{4}{5}) + 2 πn, a = - arccos (- \frac{4}{5}) + 2 πn, a = 2 πn

将解代入原方程进行验证

将它们代入

3 sin (a) + cos (a) = 1

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{4}{5}) + 2 πn

的解:

真

arccos (- \frac{4}{5}) + 2 πn

代入

n = 1

arccos (- \frac{4}{5}) + 2 π 1

对于

3 sin (a) + cos (a) = 1

代入

a = arccos (- \frac{4}{5}) + 2 π 1

3 sin (arccos (- \frac{4}{5}) + 2 π 1) + cos (arccos (- \frac{4}{5}) + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

检验

- arccos (- \frac{4}{5}) + 2 πn

的解:

假

- arccos (- \frac{4}{5}) + 2 πn

代入

n = 1

- arccos (- \frac{4}{5}) + 2 π 1

对于

3 sin (a) + cos (a) = 1

代入

a = - arccos (- \frac{4}{5}) + 2 π 1

3 sin (- arccos (- \frac{4}{5}) + 2 π 1) + cos (- arccos (- \frac{4}{5}) + 2 π 1) = 1

整理后得

- 2.6 = 1

\Rightarrow 假

检验

2 πn

的解:

真

2 πn

代入

n = 1

2 π 1

对于

3 sin (a) + cos (a) = 1

代入

a = 2 π 1

3 sin (2 π 1) + cos (2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

a = arccos (- \frac{4}{5}) + 2 πn, a = 2 πn

以小数形式表示解

a = 2.49809 \dots + 2 πn, a = 2 πn

作图

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