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受欢迎的三角函数 >

cos(x)-sin(x)= 2/3

解答

cos (x) - sin (x) = \frac{2}{3}

解答

x = π + 1.27628 \dots + 2 πn, x = 0.29451 \dots + 2 πn

+1

度数

x = 253.12550 \dots^{\circ} + 36 0^{\circ} n, x = 16.87449 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos (x) - sin (x) = \frac{2}{3}

两边加上

sin (x)

cos (x) = \frac{2}{3} + sin (x)

两边进行平方

cos^{2} (x) = (\frac{2}{3} + sin (x))^{2}

两边减去

(\frac{2}{3} + sin (x))^{2}

cos^{2} (x) - \frac{4}{9} - \frac{4}{3} sin (x) - sin^{2} (x) = 0

化简

cos^{2} (x) - \frac{4}{9} - \frac{4}{3} sin (x) - sin^{2} (x) : \frac{9 cos ^{2} ( x ) - 4 - 12 sin ( x ) - 9 sin ^{2} ( x )}{9}

cos^{2} (x) - \frac{4}{9} - \frac{4}{3} sin (x) - sin^{2} (x)

乘

\frac{4}{3} sin (x) : \frac{4 sin ( x )}{3}

\frac{4}{3} sin (x)

分式相乘:

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

= \frac{4 sin ( x )}{3}

= cos^{2} (x) - \frac{4}{9} - \frac{4 sin ( x )}{3} - sin^{2} (x)

将项转换为分式:

cos^{2} (x) = \frac{cos ^{2} ( x )}{1}, sin^{2} (x) = \frac{sin ^{2} ( x )}{1}

= \frac{cos ^{2} ( x )}{1} - \frac{4}{9} - \frac{4 sin ( x )}{3} - \frac{sin ^{2} ( x )}{1}

1, 9, 3, 1

的最小公倍数:

9

1, 9, 3, 1

最小公倍数 (LCM)

1

质因数分解

9

质因数分解:

3 \cdot 3

9

9

除以

39 = 3 \cdot 3

= 3 \cdot 3

3

质因数分解:

3

3

3

是质数，因此无法因数分解

= 3

1

质因数分解

计算出由至少在以下一个数字中出现的因数组成的数字:

1, 9, 3, 1

= 3 \cdot 3

数字相乘:

3 \cdot 3 = 9

= 9

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

9

对于

\frac{cos ^{2} ( x )}{1} :

将分母和分子乘以

9

\frac{cos ^{2} ( x )}{1} = \frac{cos ^{2} ( x ) \cdot 9}{1 \cdot 9} = \frac{cos ^{2} ( x ) \cdot 9}{9}

对于

\frac{4 sin ( x )}{3} :

将分母和分子乘以

3

\frac{4 sin ( x )}{3} = \frac{4 sin ( x ) \cdot 3}{3 \cdot 3} = \frac{12 sin ( x )}{9}

对于

\frac{sin ^{2} ( x )}{1} :

将分母和分子乘以

9

\frac{sin ^{2} ( x )}{1} = \frac{sin ^{2} ( x ) \cdot 9}{1 \cdot 9} = \frac{sin ^{2} ( x ) \cdot 9}{9}

= \frac{cos ^{2} ( x ) \cdot 9}{9} - \frac{4}{9} - \frac{12 sin ( x )}{9} - \frac{sin ^{2} ( x ) \cdot 9}{9}

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{cos ^{2} ( x ) \cdot 9 - 4 - 12 sin ( x ) - sin ^{2} ( x ) \cdot 9}{9}

\frac{9 cos ^{2} ( x ) - 4 - 12 sin ( x ) - 9 sin ^{2} ( x )}{9} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

9 cos^{2} (x) - 4 - 12 sin (x) - 9 sin^{2} (x) = 0

使用三角恒等式改写

- 4 - 12 sin (x) + 9 cos^{2} (x) - 9 sin^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 4 - 12 sin (x) + 9 (1 - sin^{2} (x)) - 9 sin^{2} (x)

化简

- 4 - 12 sin (x) + 9 (1 - sin^{2} (x)) - 9 sin^{2} (x) : - 18 sin^{2} (x) - 12 sin (x) + 5

- 4 - 12 sin (x) + 9 (1 - sin^{2} (x)) - 9 sin^{2} (x)

乘开

9 (1 - sin^{2} (x)) : 9 - 9 sin^{2} (x)

9 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

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

= 9 \cdot 1 - 9 sin^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 sin^{2} (x)

= - 4 - 12 sin (x) + 9 - 9 sin^{2} (x) - 9 sin^{2} (x)

化简

- 4 - 12 sin (x) + 9 - 9 sin^{2} (x) - 9 sin^{2} (x) : - 18 sin^{2} (x) - 12 sin (x) + 5

- 4 - 12 sin (x) + 9 - 9 sin^{2} (x) - 9 sin^{2} (x)

同类项相加:

- 9 sin^{2} (x) - 9 sin^{2} (x) = - 18 sin^{2} (x)

= - 4 - 12 sin (x) + 9 - 18 sin^{2} (x)

对同类项分组

= - 12 sin (x) - 18 sin^{2} (x) - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= - 18 sin^{2} (x) - 12 sin (x) + 5

= - 18 sin^{2} (x) - 12 sin (x) + 5

= - 18 sin^{2} (x) - 12 sin (x) + 5

5 - 12 sin (x) - 18 sin^{2} (x) = 0

用替代法求解

5 - 12 sin (x) - 18 sin^{2} (x) = 0

令

: sin (x) = u

5 - 12 u - 18 u^{2} = 0

5 - 12 u - 18 u^{2} = 0 : u = - \frac{2 + 14}{6}, u = \frac{14 - 2}{6}

5 - 12 u - 18 u^{2} = 0

改写成标准形式

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

- 18 u^{2} - 12 u + 5 = 0

使用求根公式求解

- 18 u^{2} - 12 u + 5 = 0

二次方程求根公式:

若

a = - 18, b = - 12, c = 5

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 18 ) \cdot 5}{2 ( - 18 )}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 18 ) \cdot 5}{2 ( - 18 )}

(- 12)^{2} - 4 (- 18) \cdot 5 = 614

(- 12)^{2} - 4 (- 18) \cdot 5

使用法则

- (- a) = a

= (- 12)^{2} + 4 \cdot 18 \cdot 5

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} + 4 \cdot 18 \cdot 5

数字相乘:

4 \cdot 18 \cdot 5 = 360

= 1 2^{2} + 360

1 2^{2} = 144

= 144 + 360

数字相加:

144 + 360 = 504

= 504

504

质因数分解:

2^{3} \cdot 3^{2} \cdot 7

504

504

除以

2504 = 252 \cdot 2

= 2 \cdot 252

252

除以

2252 = 126 \cdot 2

= 2 \cdot 2 \cdot 126

126

除以

2126 = 63 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 63

63

除以

363 = 21 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 21

21

除以

321 = 7 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 7

2, 3, 7

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 7

= 2^{3} \cdot 3^{2} \cdot 7

= 2^{3} \cdot 3^{2} \cdot 7

使用指数法则:

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

= 2^{2} \cdot 3^{2} \cdot 2 \cdot 7

使用根式运算法则:

n ab = n a n b

= 2^{2} 3^{2} 2 \cdot 7

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 2 \cdot 7

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 2 \cdot 7

整理后得

= 614

u_{1, 2} = \frac{- ( - 12 ) \pm 6 14}{2 ( - 18 )}

将解分隔开

u_{1} = \frac{- ( - 12 ) + 6 14}{2 ( - 18 )}, u_{2} = \frac{- ( - 12 ) - 6 14}{2 ( - 18 )}

u = \frac{- ( - 12 ) + 6 14}{2 ( - 18 )} : - \frac{2 + 14}{6}

\frac{- ( - 12 ) + 6 14}{2 ( - 18 )}

去除括号:

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

= \frac{12 + 6 14}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{12 + 6 14}{- 36}

使用分式法则:

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

= - \frac{12 + 6 14}{36}

消掉

\frac{12 + 6 14}{36} : \frac{2 + 14}{6}

\frac{12 + 6 14}{36}

分解

12 + 614 : 6 (2 + 14)

12 + 614

改写为

= 6 \cdot 2 + 614

因式分解出通项

6

= 6 (2 + 14)

= \frac{6 ( 2 + 14 )}{36}

约分:

6

= \frac{2 + 14}{6}

= - \frac{2 + 14}{6}

u = \frac{- ( - 12 ) - 6 14}{2 ( - 18 )} : \frac{14 - 2}{6}

\frac{- ( - 12 ) - 6 14}{2 ( - 18 )}

去除括号:

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

= \frac{12 - 6 14}{- 2 \cdot 18}

数字相乘:

2 \cdot 18 = 36

= \frac{12 - 6 14}{- 36}

使用分式法则:

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

12 - 614 = - (614 - 12)

= \frac{6 14 - 12}{36}

分解

614 - 12 : 6 (14 - 2)

614 - 12

改写为

= 614 - 6 \cdot 2

因式分解出通项

6

= 6 (14 - 2)

= \frac{6 ( 14 - 2 )}{36}

约分:

6

= \frac{14 - 2}{6}

二次方程组的解是:

u = - \frac{2 + 14}{6}, u = \frac{14 - 2}{6}

u = sin (x)

代回

sin (x) = - \frac{2 + 14}{6}, sin (x) = \frac{14 - 2}{6}

sin (x) = - \frac{2 + 14}{6}, sin (x) = \frac{14 - 2}{6}

sin (x) = - \frac{2 + 14}{6} : x = arcsin (- \frac{2 + 14}{6}) + 2 πn, x = π + arcsin (\frac{2 + 14}{6}) + 2 πn

sin (x) = - \frac{2 + 14}{6}

使用反三角函数性质

sin (x) = - \frac{2 + 14}{6}

sin (x) = - \frac{2 + 14}{6}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{2 + 14}{6}) + 2 πn, x = π + arcsin (\frac{2 + 14}{6}) + 2 πn

x = arcsin (- \frac{2 + 14}{6}) + 2 πn, x = π + arcsin (\frac{2 + 14}{6}) + 2 πn

sin (x) = \frac{14 - 2}{6} : x = arcsin (\frac{14 - 2}{6}) + 2 πn, x = π - arcsin (\frac{14 - 2}{6}) + 2 πn

sin (x) = \frac{14 - 2}{6}

使用反三角函数性质

sin (x) = \frac{14 - 2}{6}

sin (x) = \frac{14 - 2}{6}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{14 - 2}{6}) + 2 πn, x = π - arcsin (\frac{14 - 2}{6}) + 2 πn

x = arcsin (\frac{14 - 2}{6}) + 2 πn, x = π - arcsin (\frac{14 - 2}{6}) + 2 πn

合并所有解

x = arcsin (- \frac{2 + 14}{6}) + 2 πn, x = π + arcsin (\frac{2 + 14}{6}) + 2 πn, x = arcsin (\frac{14 - 2}{6}) + 2 πn, x = π - arcsin (\frac{14 - 2}{6}) + 2 πn

将解代入原方程进行验证

将它们代入

cos (x) - sin (x) = \frac{2}{3}

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{2 + 14}{6}) + 2 πn

的解:

假

arcsin (- \frac{2 + 14}{6}) + 2 πn

代入

n = 1

arcsin (- \frac{2 + 14}{6}) + 2 π 1

对于

cos (x) - sin (x) = \frac{2}{3}

代入

x = arcsin (- \frac{2 + 14}{6}) + 2 π 1

cos (arcsin (- \frac{2 + 14}{6}) + 2 π 1) - sin (arcsin (- \frac{2 + 14}{6}) + 2 π 1) = \frac{2}{3}

整理后得

1.24721 \dots = 0.66666 \dots

\Rightarrow 假

检验

π + arcsin (\frac{2 + 14}{6}) + 2 πn

的解:

真

π + arcsin (\frac{2 + 14}{6}) + 2 πn

代入

n = 1

π + arcsin (\frac{2 + 14}{6}) + 2 π 1

对于

cos (x) - sin (x) = \frac{2}{3}

代入

x = π + arcsin (\frac{2 + 14}{6}) + 2 π 1

cos (π + arcsin (\frac{2 + 14}{6}) + 2 π 1) - sin (π + arcsin (\frac{2 + 14}{6}) + 2 π 1) = \frac{2}{3}

整理后得

0.66666 \dots = 0.66666 \dots

\Rightarrow 真

检验

arcsin (\frac{14 - 2}{6}) + 2 πn

的解:

真

arcsin (\frac{14 - 2}{6}) + 2 πn

代入

n = 1

arcsin (\frac{14 - 2}{6}) + 2 π 1

对于

cos (x) - sin (x) = \frac{2}{3}

代入

x = arcsin (\frac{14 - 2}{6}) + 2 π 1

cos (arcsin (\frac{14 - 2}{6}) + 2 π 1) - sin (arcsin (\frac{14 - 2}{6}) + 2 π 1) = \frac{2}{3}

整理后得

0.66666 \dots = 0.66666 \dots

\Rightarrow 真

检验

π - arcsin (\frac{14 - 2}{6}) + 2 πn

的解:

假

π - arcsin (\frac{14 - 2}{6}) + 2 πn

代入

n = 1

π - arcsin (\frac{14 - 2}{6}) + 2 π 1

对于

cos (x) - sin (x) = \frac{2}{3}

代入

x = π - arcsin (\frac{14 - 2}{6}) + 2 π 1

cos (π - arcsin (\frac{14 - 2}{6}) + 2 π 1) - sin (π - arcsin (\frac{14 - 2}{6}) + 2 π 1) = \frac{2}{3}

整理后得

- 1.24721 \dots = 0.66666 \dots

\Rightarrow 假

x = π + arcsin (\frac{2 + 14}{6}) + 2 πn, x = arcsin (\frac{14 - 2}{6}) + 2 πn

以小数形式表示解

x = π + 1.27628 \dots + 2 πn, x = 0.29451 \dots + 2 πn

作图

流行的例子

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arctan(x+1/3)+arctan(x-1/3)=arctan(2)

arctan (x + \frac{1}{3}) + arctan (x - \frac{1}{3}) = arctan (2)

sin (θ) = 0, 64

cos(3x)=cos(2x)

cos (3 x) = cos (2 x)

2cos(x)cos^3(x)+cos^2(x)-sin^2(x)=cos(x)

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