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受欢迎的三角函数 >

3sqrt(2)sin(v)+3sqrt(2)cos(v)=3

解答

32 sin (v) + 32 cos (v) = 3

解答

v = 1.83259 \dots + 2 πn, v = 2 π - 0.26179 \dots + 2 πn

+1

度数

v = 10 5^{\circ} + 36 0^{\circ} n, v = 34 5^{\circ} + 36 0^{\circ} n

求解步骤

32 sin (v) + 32 cos (v) = 3

两边减去

32 cos (v)

32 sin (v) = 3 - 32 cos (v)

两边进行平方

(32 sin (v))^{2} = (3 - 32 cos (v))^{2}

两边减去

(3 - 32 cos (v))^{2}

18 sin^{2} (v) - 9 + 182 cos (v) - 18 cos^{2} (v) = 0

使用三角恒等式改写

- 9 - 18 cos^{2} (v) + 18 sin^{2} (v) + 18 cos (v) 2

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 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2

化简

- 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2 : 182 cos (v) - 36 cos^{2} (v) + 9

- 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 18 cos (v) 2

= - 9 - 18 cos^{2} (v) + 18 (1 - cos^{2} (v)) + 182 cos (v)

乘开

18 (1 - cos^{2} (v)) : 18 - 18 cos^{2} (v)

18 (1 - cos^{2} (v))

使用分配律:

a (b - c) = ab - a c

a = 18, b = 1, c = cos^{2} (v)

= 18 \cdot 1 - 18 cos^{2} (v)

数字相乘:

18 \cdot 1 = 18

= 18 - 18 cos^{2} (v)

= - 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2

化简

- 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2 : 182 cos (v) - 36 cos^{2} (v) + 9

- 9 - 18 cos^{2} (v) + 18 - 18 cos^{2} (v) + 18 cos (v) 2

对同类项分组

= - 18 cos^{2} (v) - 18 cos^{2} (v) + 182 cos (v) - 9 + 18

同类项相加:

- 18 cos^{2} (v) - 18 cos^{2} (v) = - 36 cos^{2} (v)

= - 36 cos^{2} (v) + 182 cos (v) - 9 + 18

数字相加/相减:

- 9 + 18 = 9

= 182 cos (v) - 36 cos^{2} (v) + 9

= 182 cos (v) - 36 cos^{2} (v) + 9

= 182 cos (v) - 36 cos^{2} (v) + 9

9 - 36 cos^{2} (v) + 18 cos (v) 2 = 0

用替代法求解

9 - 36 cos^{2} (v) + 18 cos (v) 2 = 0

令

: cos (v) = u

9 - 36 u^{2} + 18 u 2 = 0

9 - 36 u^{2} + 18 u 2 = 0 : u = - \frac{- 2 + 6}{4}, u = \frac{2 + 6}{4}

9 - 36 u^{2} + 18 u 2 = 0

改写成标准形式

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

- 36 u^{2} + 182 u + 9 = 0

使用求根公式求解

- 36 u^{2} + 182 u + 9 = 0

二次方程求根公式:

若

a = - 36, b = 182, c = 9

u_{1, 2} = \frac{- 18 2 \pm ( 18 2 ) ^{2} - 4 ( - 36 ) \cdot 9}{2 ( - 36 )}

u_{1, 2} = \frac{- 18 2 \pm ( 18 2 ) ^{2} - 4 ( - 36 ) \cdot 9}{2 ( - 36 )}

(182)^{2} - 4 (- 36) \cdot 9 = 186

(182)^{2} - 4 (- 36) \cdot 9

使用法则

- (- a) = a

= (182)^{2} + 4 \cdot 36 \cdot 9

(182)^{2} = 1 8^{2} \cdot 2

(182)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 1 8^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

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

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 1 8^{2} \cdot 2

4 \cdot 36 \cdot 9 = 1296

4 \cdot 36 \cdot 9

数字相乘:

4 \cdot 36 \cdot 9 = 1296

= 1296

= 1 8^{2} \cdot 2 + 1296

1 8^{2} \cdot 2 = 648

1 8^{2} \cdot 2

1 8^{2} = 324

= 324 \cdot 2

数字相乘:

324 \cdot 2 = 648

= 648

= 648 + 1296

数字相加:

648 + 1296 = 1944

= 1944

1944

质因数分解:

2^{3} \cdot 3^{5}

1944

1944

除以

21944 = 972 \cdot 2

= 2 \cdot 972

972

除以

2972 = 486 \cdot 2

= 2 \cdot 2 \cdot 486

486

除以

2486 = 243 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 243

243

除以

3243 = 81 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 81

81

除以

381 = 27 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 27

27

除以

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{5}

= 3^{5} \cdot 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 3^{4} \cdot 2^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 3^{4} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{4} 2 \cdot 3

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

3^{4} = 3^{\frac{4}{2}} = 3^{2}

= 3^{2} \cdot 2 2 \cdot 3

整理后得

= 186

u_{1, 2} = \frac{- 18 2 \pm 18 6}{2 ( - 36 )}

将解分隔开

u_{1} = \frac{- 18 2 + 18 6}{2 ( - 36 )}, u_{2} = \frac{- 18 2 - 18 6}{2 ( - 36 )}

u = \frac{- 18 2 + 18 6}{2 ( - 36 )} : - \frac{- 2 + 6}{4}

\frac{- 18 2 + 18 6}{2 ( - 36 )}

去除括号:

(- a) = - a

= \frac{- 18 2 + 18 6}{- 2 \cdot 36}

数字相乘:

2 \cdot 36 = 72

= \frac{- 18 2 + 18 6}{- 72}

使用分式法则:

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

= - \frac{- 18 2 + 18 6}{72}

消掉

\frac{- 18 2 + 18 6}{72} : \frac{6 - 2}{4}

\frac{- 18 2 + 18 6}{72}

因式分解出通项

18

= \frac{18 ( - 2 + 6 )}{72}

约分:

18

= \frac{- 2 + 6}{4}

= - \frac{6 - 2}{4}

= - \frac{- 2 + 6}{4}

u = \frac{- 18 2 - 18 6}{2 ( - 36 )} : \frac{2 + 6}{4}

\frac{- 18 2 - 18 6}{2 ( - 36 )}

去除括号:

(- a) = - a

= \frac{- 18 2 - 18 6}{- 2 \cdot 36}

数字相乘:

2 \cdot 36 = 72

= \frac{- 18 2 - 18 6}{- 72}

使用分式法则:

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

- 182 - 186 = - (182 + 186)

= \frac{18 2 + 18 6}{72}

因式分解出通项

18

= \frac{18 ( 2 + 6 )}{72}

约分:

18

= \frac{2 + 6}{4}

二次方程组的解是:

u = - \frac{- 2 + 6}{4}, u = \frac{2 + 6}{4}

u = cos (v)

代回

cos (v) = - \frac{- 2 + 6}{4}, cos (v) = \frac{2 + 6}{4}

cos (v) = - \frac{- 2 + 6}{4}, cos (v) = \frac{2 + 6}{4}

cos (v) = - \frac{- 2 + 6}{4} : v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

cos (v) = - \frac{- 2 + 6}{4}

使用反三角函数性质

cos (v) = - \frac{- 2 + 6}{4}

cos (v) = - \frac{- 2 + 6}{4}

的通解

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

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn

cos (v) = \frac{2 + 6}{4} : v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

cos (v) = \frac{2 + 6}{4}

使用反三角函数性质

cos (v) = \frac{2 + 6}{4}

cos (v) = \frac{2 + 6}{4}

的通解

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

v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

合并所有解

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = - arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = arccos (\frac{2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

将解代入原方程进行验证

将它们代入

32 sin (v) + 32 cos (v) = 3

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{- 2 + 6}{4}) + 2 πn

的解:

真

arccos (- \frac{- 2 + 6}{4}) + 2 πn

代入

n = 1

arccos (- \frac{- 2 + 6}{4}) + 2 π 1

对于

32 sin (v) + 32 cos (v) = 3

代入

v = arccos (- \frac{- 2 + 6}{4}) + 2 π 1

32 sin (arccos (- \frac{- 2 + 6}{4}) + 2 π 1) + 32 cos (arccos (- \frac{- 2 + 6}{4}) + 2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

检验

- arccos (- \frac{- 2 + 6}{4}) + 2 πn

的解:

假

- arccos (- \frac{- 2 + 6}{4}) + 2 πn

代入

n = 1

- arccos (- \frac{- 2 + 6}{4}) + 2 π 1

对于

32 sin (v) + 32 cos (v) = 3

代入

v = - arccos (- \frac{- 2 + 6}{4}) + 2 π 1

32 sin (- arccos (- \frac{- 2 + 6}{4}) + 2 π 1) + 32 cos (- arccos (- \frac{- 2 + 6}{4}) + 2 π 1) = 3

整理后得

- 5.19615 \dots = 3

\Rightarrow 假

检验

arccos (\frac{2 + 6}{4}) + 2 πn

的解:

假

arccos (\frac{2 + 6}{4}) + 2 πn

代入

n = 1

arccos (\frac{2 + 6}{4}) + 2 π 1

对于

32 sin (v) + 32 cos (v) = 3

代入

v = arccos (\frac{2 + 6}{4}) + 2 π 1

32 sin (arccos (\frac{2 + 6}{4}) + 2 π 1) + 32 cos (arccos (\frac{2 + 6}{4}) + 2 π 1) = 3

整理后得

5.19615 \dots = 3

\Rightarrow 假

检验

2 π - arccos (\frac{2 + 6}{4}) + 2 πn

的解:

真

2 π - arccos (\frac{2 + 6}{4}) + 2 πn

代入

n = 1

2 π - arccos (\frac{2 + 6}{4}) + 2 π 1

对于

32 sin (v) + 32 cos (v) = 3

代入

v = 2 π - arccos (\frac{2 + 6}{4}) + 2 π 1

32 sin (2 π - arccos (\frac{2 + 6}{4}) + 2 π 1) + 32 cos (2 π - arccos (\frac{2 + 6}{4}) + 2 π 1) = 3

整理后得

3 = 3

\Rightarrow 真

v = arccos (- \frac{- 2 + 6}{4}) + 2 πn, v = 2 π - arccos (\frac{2 + 6}{4}) + 2 πn

以小数形式表示解

v = 1.83259 \dots + 2 πn, v = 2 π - 0.26179 \dots + 2 πn

作图

流行的例子

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