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受欢迎的三角函数 >

tan(arcsin(2/3)-arccos(1/3))

解答

tan (arcsin (\frac{2}{3}) - arccos (\frac{1}{3}))

解答

- \frac{2 ( 5 - 2 )}{3}

+1

十进制

- 0.54790 \dots

求解步骤

tan (arcsin (\frac{2}{3}) - arccos (\frac{1}{3}))

使用三角恒等式改写:

\frac{tan ( arcsin ( \frac{2}{3} ) ) - tan ( arccos ( \frac{1}{3} ) )}{1 + tan ( arcsin ( \frac{2}{3} ) ) tan ( arccos ( \frac{1}{3} ) )}

tan (arcsin (\frac{2}{3}) - arccos (\frac{1}{3}))

使用角差恒等式:

tan (s - t) = \frac{tan ( s ) - tan ( t )}{1 + tan ( s ) tan ( t )}

= \frac{tan ( arcsin ( \frac{2}{3} ) ) - tan ( arccos ( \frac{1}{3} ) )}{1 + tan ( arcsin ( \frac{2}{3} ) ) tan ( arccos ( \frac{1}{3} ) )}

= \frac{tan ( arcsin ( \frac{2}{3} ) ) - tan ( arccos ( \frac{1}{3} ) )}{1 + tan ( arcsin ( \frac{2}{3} ) ) tan ( arccos ( \frac{1}{3} ) )}

使用三角恒等式改写:

tan (arcsin (\frac{2}{3})) = \frac{2 5}{5}

tan (arcsin (\frac{2}{3}))

使用三角恒等式改写:

tan (arcsin (\frac{2}{3})) = \frac{( \frac{2}{3} ) 1 - ( \frac{2}{3} ) ^{2}}{1 - ( \frac{2}{3} ) ^{2}}

利用以下特性:

tan (arcsin (x)) = \frac{x 1 - x ^{2}}{1 - x ^{2}}

= \frac{( \frac{2}{3} ) 1 - ( \frac{2}{3} ) ^{2}}{1 - ( \frac{2}{3} ) ^{2}}

= \frac{\frac{2}{3} 1 - ( \frac{2}{3} ) ^{2}}{1 - ( \frac{2}{3} ) ^{2}}

化简

= \frac{2 5}{5}

使用三角恒等式改写:

tan (arccos (\frac{1}{3})) = 22

tan (arccos (\frac{1}{3}))

使用三角恒等式改写:

tan (arccos (\frac{1}{3})) = \frac{1 - ( \frac{1}{3} ) ^{2}}{( \frac{1}{3} )}

利用以下特性:

tan (arccos (x)) = \frac{1 - x ^{2}}{x}

= \frac{1 - ( \frac{1}{3} ) ^{2}}{( \frac{1}{3} )}

= \frac{1 - ( \frac{1}{3} ) ^{2}}{\frac{1}{3}}

化简

= 22

= \frac{\frac{2 5}{5} - 2 2}{1 + \frac{2 5}{5} \cdot 2 2}

化简

\frac{\frac{2 5}{5} - 2 2}{1 + \frac{2 5}{5} \cdot 2 2} : - \frac{2 ( 5 - 2 )}{3}

\frac{\frac{2 5}{5} - 2 2}{1 + \frac{2 5}{5} \cdot 2 2}

\frac{2 5}{5} \cdot 22 = \frac{4 10}{5}

\frac{2 5}{5} \cdot 22

分式相乘:

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

= \frac{2 5 \cdot 2 2}{5}

数字相乘:

2 \cdot 2 = 4

= \frac{4 5 2}{5}

化简

452 : 410

452

使用根式运算法则:

a b = a \cdot b

52 = 5 \cdot 2

= 4 5 \cdot 2

数字相乘:

5 \cdot 2 = 10

= 410

= \frac{4 10}{5}

= \frac{\frac{2 5}{5} - 2 2}{1 + \frac{4 10}{5}}

化简

\frac{2 5}{5} - 22 : \frac{2 5 - 10 2}{5}

\frac{2 5}{5} - 22

将项转换为分式:

22 = \frac{2 \cdot 2 \cdot 5}{5}

= \frac{2 5}{5} - \frac{2 2 \cdot 5}{5}

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{2 5 - 2 2 \cdot 5}{5}

数字相乘:

2 \cdot 5 = 10

= \frac{2 5 - 10 2}{5}

= \frac{\frac{2 5 - 10 2}{5}}{1 + \frac{4 10}{5}}

使用分式法则:

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

= \frac{2 5 - 10 2}{5 ( 1 + \frac{4 10}{5} )}

化简

1 + \frac{4 10}{5} : \frac{5 + 4 10}{5}

1 + \frac{4 10}{5}

将项转换为分式:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} + \frac{4 10}{5}

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{1 \cdot 5 + 4 10}{5}

数字相乘:

1 \cdot 5 = 5

= \frac{5 + 4 10}{5}

= \frac{2 5 - 10 2}{5 \cdot \frac{5 + 4 10}{5}}

乘

5 \cdot \frac{5 + 4 10}{5} : 5 + 410

5 \cdot \frac{5 + 4 10}{5}

分式相乘:

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

= \frac{( 5 + 4 10 ) \cdot 5}{5}

约分:

5

= 5 + 410

= \frac{2 5 - 10 2}{5 + 4 10}

\frac{2 5 - 10 2}{5 + 4 10}

有理化:

- \frac{2 ( 5 - 2 )}{3}

\frac{2 5 - 10 2}{5 + 4 10}

乘以共轭根式

\frac{5 - 4 10}{5 - 4 10}

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

(25 - 102) (5 - 410) = 905 - 902

(25 - 102) (5 - 410)

使用 FOIL 方法:

(a + b) (c + d) = a c + a d + b c + b d

a = 25, b = - 102, c = 5, d = - 410

= 25 \cdot 5 + 25 (- 410) + (- 102) \cdot 5 + (- 102) (- 410)

使用加减运算法则

+ (- a) = - a, (- a) (- b) = ab

= 2 \cdot 55 - 2 \cdot 4510 - 10 \cdot 52 + 10 \cdot 4210

化简

2 \cdot 55 - 2 \cdot 4510 - 10 \cdot 52 + 10 \cdot 4210 : 905 - 902

2 \cdot 55 - 2 \cdot 4510 - 10 \cdot 52 + 10 \cdot 4210

2 \cdot 55 = 105

2 \cdot 55

数字相乘:

2 \cdot 5 = 10

= 105

2 \cdot 4510 = 402

2 \cdot 4510

分解整数

10 = 5 \cdot 2

= 2 \cdot 45 5 \cdot 2

使用根式运算法则:

n ab = n a n b

5 \cdot 2 = 52

= 2 \cdot 4552

使用根式运算法则:

a a = a

55 = 5

= 2 \cdot 4 \cdot 52

数字相乘:

2 \cdot 4 \cdot 5 = 40

= 402

10 \cdot 52 = 502

10 \cdot 52

数字相乘:

10 \cdot 5 = 50

= 502

10 \cdot 4210 = 805

10 \cdot 4210

分解整数

10 = 2 \cdot 5

= 2 \cdot 5 \cdot 4210

分解整数

4 = 2^{2}

= 2 \cdot 5 \cdot 2^{2} 210

分解整数

10 = 2 \cdot 5

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

使用根式运算法则:

n ab = n a n b

2 \cdot 5 = 25

= 2 \cdot 5 \cdot 2^{2} 225

使用指数法则:

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

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

= 5 \cdot 2^{1 + 2} 225

数字相加:

1 + 2 = 3

= 5 \cdot 2^{3} 225

使用根式运算法则:

a a = a

22 = 2

= 2^{3} \cdot 5 \cdot 25

数字相乘:

5 \cdot 2 = 10

= 2^{3} \cdot 105

2^{3} = 8

= 10 \cdot 85

数字相乘:

10 \cdot 8 = 80

= 805

= 105 - 402 - 502 + 805

同类项相加:

- 402 - 502 = - 902

= 105 - 902 + 805

同类项相加:

105 + 805 = 905

= 905 - 902

= 905 - 902

(5 + 410) (5 - 410) = - 135

(5 + 410) (5 - 410)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 5, b = 410

= 5^{2} - (410)^{2}

化简

5^{2} - (410)^{2} : - 135

5^{2} - (410)^{2}

5^{2} = 25

5^{2}

5^{2} = 25

= 25

(410)^{2} = 160

(410)^{2}

使用指数法则:

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

= 4^{2} (10)^{2}

(10)^{2} : 10

使用根式运算法则:

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

= (1 0^{\frac{1}{2}})^{2}

使用指数法则:

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

= 1 0^{\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

= 10

= 4^{2} \cdot 10

4^{2} = 16

= 16 \cdot 10

数字相乘:

16 \cdot 10 = 160

= 160

= 25 - 160

数字相减:

25 - 160 = - 135

= - 135

= - 135

= \frac{90 5 - 90 2}{- 135}

使用分式法则:

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

= - \frac{90 5 - 90 2}{135}

消掉

\frac{90 5 - 90 2}{135} : \frac{2 ( 5 - 2 )}{3}

\frac{90 5 - 90 2}{135}

因式分解出通项

90

= \frac{90 ( 5 - 2 )}{135}

约分:

45

= \frac{2 ( 5 - 2 )}{3}

= - \frac{2 ( 5 - 2 )}{3}

= - \frac{2 ( 5 - 2 )}{3}

= - \frac{2 ( 5 - 2 )}{3}

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