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受欢迎的三角函数 >

证明 cos((7pi)/(12))=cos(pi/3+pi/4)

解答

证明

cos (\frac{7 π}{12}) = cos (\frac{π}{3} + \frac{π}{4})

解答

真

求解步骤

cos (\frac{7 π}{12}) = cos (\frac{π}{3} + \frac{π}{4})

调整左侧

cos (\frac{7 π}{12})

使用三角恒等式改写

cos (\frac{7 π}{12})

= cos (2 \cdot \frac{7 π}{24})

使用倍角公式:

cos (2 \frac{7 π}{24}) = cos^{2} (\frac{7 π}{24}) - sin^{2} (\frac{7 π}{24})

= cos (\frac{7 π}{12})

分解

cos^{2} (\frac{7 π}{24}) - sin^{2} (\frac{7 π}{24}) : (cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

cos^{2} (\frac{7 π}{24}) - sin^{2} (\frac{7 π}{24})

使用平方差公式:

\frac{7 π}{24}^{2} - y^{2} = (\frac{7 π}{24} + y) (\frac{7 π}{24} - y)

cos^{2} (\frac{7 π}{24}) - sin^{2} (\frac{7 π}{24}) = (cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

= (cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

= (cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

(cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24})) = \frac{2 - 6}{4}

(cos (\frac{7 π}{24}) + sin (\frac{7 π}{24})) (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

cos (\frac{7 π}{24}) = \frac{4 + 2 ( 1 - 3 )}{2 2}

cos (\frac{7 π}{24})

使用三角恒等式改写:

\frac{1 + cos ( \frac{7 π}{12} )}{2}

cos (\frac{7 π}{24})

将

cos (\frac{7 π}{24})

写为

cos (\frac{\frac{7 π}{12}}{2})

= cos (\frac{\frac{7 π}{12}}{2})

使用半角公式:

cos (\frac{θ}{2}) = \frac{1 + cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

= \frac{1 + cos ( \frac{7 π}{12} )}{2}

= \frac{1 + cos ( \frac{7 π}{12} )}{2}

使用三角恒等式改写:

cos (\frac{7 π}{12}) = \frac{2 ( 1 - 3 )}{4}

cos (\frac{7 π}{12})

使用三角恒等式改写:

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{7 π}{12})

将

cos (\frac{7 π}{12})

写为

cos (\frac{π}{3} + \frac{π}{4})

= cos (\frac{π}{3} + \frac{π}{4})

使用角和恒等式:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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}

= \frac{2}{2}

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

化简

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

因式分解出通项

\frac{2}{2}

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

\frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}

\frac{1}{2} - \frac{3}{2}

使用法则

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

= \frac{1 - 3}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

化简

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2} : \frac{2 4 + 2 ( 1 - 3 )}{4}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2} = \frac{4 + 2 ( 1 - 3 )}{8}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

化简

1 + \frac{2 ( 1 - 3 )}{4} : \frac{4 + 2 ( 1 - 3 )}{4}

1 + \frac{2 ( 1 - 3 )}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{2 ( 1 - 3 )}{4}

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 4 + 2 ( 1 - 3 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 + 2 ( 1 - 3 )}{4}

= \frac{\frac{4 + 2 ( 1 - 3 )}{4}}{2}

使用分式法则:

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

= \frac{4 + 2 ( 1 - 3 )}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{4 + 2 ( 1 - 3 )}{8}

= \frac{4 + 2 ( 1 - 3 )}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4 + 2 ( 1 - 3 )}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{2 ( 1 - 3 ) + 4}{2 2}

\frac{4 + 2 ( 1 - 3 )}{2 2}

有理化:

\frac{2 2 ( 1 - 3 ) + 4}{4}

\frac{4 + 2 ( 1 - 3 )}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{4 + 2 ( 1 - 3 ) 2}{2 2 2}

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

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

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 4 + 2 ( 1 - 3 )}{4}

= \frac{2 2 ( 1 - 3 ) + 4}{4}

= \frac{2 4 + 2 ( 1 - 3 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 2 ( 1 - 3 ) + 4}{2 ^{2}}

消掉

\frac{2 4 + 2 ( 1 - 3 )}{2 ^{2}} : \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

\frac{2 4 + 2 ( 1 - 3 )}{2 ^{2}}

使用根式运算法则:

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

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

= \frac{2 ^{\frac{1}{2}} 2 ( 1 - 3 ) + 4}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{4 + 2 ( 1 - 3 )}{2 ^{2 - \frac{1}{2}}}

数字相减:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

= \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{4 + 2 ( 1 - 3 )}{2 2}

sin (\frac{7 π}{24}) = \frac{4 - 2 ( 1 - 3 )}{2 2}

sin (\frac{7 π}{24})

使用三角恒等式改写:

\frac{1 - cos ( \frac{7 π}{12} )}{2}

sin (\frac{7 π}{24})

将

sin (\frac{7 π}{24})

写为

sin (\frac{\frac{7 π}{12}}{2})

= sin (\frac{\frac{7 π}{12}}{2})

使用半角公式:

sin (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

用

\frac{θ}{2}

替代

θ

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

交换两边

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

两边除以

2

sin^{2} (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

sin (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

= \frac{1 - cos ( \frac{7 π}{12} )}{2}

= \frac{1 - cos ( \frac{7 π}{12} )}{2}

使用三角恒等式改写:

cos (\frac{7 π}{12}) = \frac{2 ( 1 - 3 )}{4}

cos (\frac{7 π}{12})

使用三角恒等式改写:

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{7 π}{12})

将

cos (\frac{7 π}{12})

写为

cos (\frac{π}{3} + \frac{π}{4})

= cos (\frac{π}{3} + \frac{π}{4})

使用角和恒等式:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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}

= \frac{2}{2}

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

化简

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

因式分解出通项

\frac{2}{2}

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

\frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}

\frac{1}{2} - \frac{3}{2}

使用法则

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

= \frac{1 - 3}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

化简

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2} : \frac{2 4 - 2 ( 1 - 3 )}{4}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2} = \frac{4 - 2 ( 1 - 3 )}{8}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

化简

1 - \frac{2 ( 1 - 3 )}{4} : \frac{4 - 2 ( 1 - 3 )}{4}

1 - \frac{2 ( 1 - 3 )}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{2 ( 1 - 3 )}{4}

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 4 - 2 ( 1 - 3 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 - 2 ( 1 - 3 )}{4}

= \frac{\frac{4 - 2 ( 1 - 3 )}{4}}{2}

使用分式法则:

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

= \frac{4 - 2 ( 1 - 3 )}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{4 - 2 ( 1 - 3 )}{8}

= \frac{4 - 2 ( 1 - 3 )}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4 - 2 ( 1 - 3 )}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{- 2 ( 1 - 3 ) + 4}{2 2}

\frac{4 - 2 ( 1 - 3 )}{2 2}

有理化:

\frac{2 - 2 ( 1 - 3 ) + 4}{4}

\frac{4 - 2 ( 1 - 3 )}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{4 - 2 ( 1 - 3 ) 2}{2 2 2}

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

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

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 4 - 2 ( 1 - 3 )}{4}

= \frac{2 - 2 ( 1 - 3 ) + 4}{4}

= \frac{2 4 - 2 ( 1 - 3 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 - 2 ( 1 - 3 ) + 4}{2 ^{2}}

消掉

\frac{2 4 - 2 ( 1 - 3 )}{2 ^{2}} : \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

\frac{2 4 - 2 ( 1 - 3 )}{2 ^{2}}

使用根式运算法则:

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

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

= \frac{2 ^{\frac{1}{2}} - 2 ( 1 - 3 ) + 4}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{4 - 2 ( 1 - 3 )}{2 ^{2 - \frac{1}{2}}}

数字相减:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

= \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{4 - 2 ( 1 - 3 )}{2 2}

= \frac{2 ( 1 - 3 ) + 4}{2 2} + \frac{- 2 ( 1 - 3 ) + 4}{2 2} (cos (\frac{7 π}{24}) - sin (\frac{7 π}{24}))

cos (\frac{7 π}{24}) = \frac{4 + 2 ( 1 - 3 )}{2 2}

cos (\frac{7 π}{24})

使用三角恒等式改写:

\frac{1 + cos ( \frac{7 π}{12} )}{2}

cos (\frac{7 π}{24})

将

cos (\frac{7 π}{24})

写为

cos (\frac{\frac{7 π}{12}}{2})

= cos (\frac{\frac{7 π}{12}}{2})

使用半角公式:

cos (\frac{θ}{2}) = \frac{1 + cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

用

\frac{θ}{2}

替代

θ

cos (θ) = 2 cos^{2} (\frac{θ}{2}) - 1

交换两边

2 cos^{2} (\frac{θ}{2}) = 1 + cos (θ)

两边除以

2

cos^{2} (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

cos (\frac{θ}{2}) = \frac{( 1 + cos ( θ ) )}{2}

= \frac{1 + cos ( \frac{7 π}{12} )}{2}

= \frac{1 + cos ( \frac{7 π}{12} )}{2}

使用三角恒等式改写:

cos (\frac{7 π}{12}) = \frac{2 ( 1 - 3 )}{4}

cos (\frac{7 π}{12})

使用三角恒等式改写:

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{7 π}{12})

将

cos (\frac{7 π}{12})

写为

cos (\frac{π}{3} + \frac{π}{4})

= cos (\frac{π}{3} + \frac{π}{4})

使用角和恒等式:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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}

= \frac{2}{2}

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

化简

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

因式分解出通项

\frac{2}{2}

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

\frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}

\frac{1}{2} - \frac{3}{2}

使用法则

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

= \frac{1 - 3}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

化简

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2} : \frac{2 4 + 2 ( 1 - 3 )}{4}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2} = \frac{4 + 2 ( 1 - 3 )}{8}

\frac{1 + \frac{2 ( 1 - 3 )}{4}}{2}

化简

1 + \frac{2 ( 1 - 3 )}{4} : \frac{4 + 2 ( 1 - 3 )}{4}

1 + \frac{2 ( 1 - 3 )}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} + \frac{2 ( 1 - 3 )}{4}

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 4 + 2 ( 1 - 3 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 + 2 ( 1 - 3 )}{4}

= \frac{\frac{4 + 2 ( 1 - 3 )}{4}}{2}

使用分式法则:

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

= \frac{4 + 2 ( 1 - 3 )}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{4 + 2 ( 1 - 3 )}{8}

= \frac{4 + 2 ( 1 - 3 )}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4 + 2 ( 1 - 3 )}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{2 ( 1 - 3 ) + 4}{2 2}

\frac{4 + 2 ( 1 - 3 )}{2 2}

有理化:

\frac{2 2 ( 1 - 3 ) + 4}{4}

\frac{4 + 2 ( 1 - 3 )}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{4 + 2 ( 1 - 3 ) 2}{2 2 2}

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

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

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 4 + 2 ( 1 - 3 )}{4}

= \frac{2 2 ( 1 - 3 ) + 4}{4}

= \frac{2 4 + 2 ( 1 - 3 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 2 ( 1 - 3 ) + 4}{2 ^{2}}

消掉

\frac{2 4 + 2 ( 1 - 3 )}{2 ^{2}} : \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

\frac{2 4 + 2 ( 1 - 3 )}{2 ^{2}}

使用根式运算法则:

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

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

= \frac{2 ^{\frac{1}{2}} 2 ( 1 - 3 ) + 4}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{4 + 2 ( 1 - 3 )}{2 ^{2 - \frac{1}{2}}}

数字相减:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

= \frac{4 + 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{4 + 2 ( 1 - 3 )}{2 2}

sin (\frac{7 π}{24}) = \frac{4 - 2 ( 1 - 3 )}{2 2}

sin (\frac{7 π}{24})

使用三角恒等式改写:

\frac{1 - cos ( \frac{7 π}{12} )}{2}

sin (\frac{7 π}{24})

将

sin (\frac{7 π}{24})

写为

sin (\frac{\frac{7 π}{12}}{2})

= sin (\frac{\frac{7 π}{12}}{2})

使用半角公式:

sin (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

用

\frac{θ}{2}

替代

θ

cos (θ) = 1 - 2 sin^{2} (\frac{θ}{2})

交换两边

2 sin^{2} (\frac{θ}{2}) = 1 - cos (θ)

两边除以

2

sin^{2} (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

Square root both sides

Choose the root sign according to the quadrant of

\frac{θ}{2}

range [0, \frac{π}{2}] [\frac{π}{2}, π] [π, \frac{3 π}{2}] [\frac{3 π}{2}, 2 π] quadrant I II III I V sin positive positive negative negative cos positive negative negative positive

sin (\frac{θ}{2}) = \frac{( 1 - cos ( θ ) )}{2}

= \frac{1 - cos ( \frac{7 π}{12} )}{2}

= \frac{1 - cos ( \frac{7 π}{12} )}{2}

使用三角恒等式改写:

cos (\frac{7 π}{12}) = \frac{2 ( 1 - 3 )}{4}

cos (\frac{7 π}{12})

使用三角恒等式改写:

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{7 π}{12})

将

cos (\frac{7 π}{12})

写为

cos (\frac{π}{3} + \frac{π}{4})

= cos (\frac{π}{3} + \frac{π}{4})

使用角和恒等式:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

cos (\frac{π}{3})

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}

= \frac{1}{2}

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (\frac{π}{4})

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}

= \frac{2}{2}

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (\frac{π}{3})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (\frac{π}{4})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

化简

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} : \frac{2 ( 1 - 3 )}{4}

\frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2}

因式分解出通项

\frac{2}{2}

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

\frac{1}{2} - \frac{3}{2} = \frac{1 - 3}{2}

\frac{1}{2} - \frac{3}{2}

使用法则

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

= \frac{1 - 3}{2}

= \frac{2}{2} \cdot \frac{1 - 3}{2}

分式相乘:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 ( 1 - 3 )}{4}

= \frac{2 ( 1 - 3 )}{4}

= \frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

化简

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2} : \frac{2 4 - 2 ( 1 - 3 )}{4}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2} = \frac{4 - 2 ( 1 - 3 )}{8}

\frac{1 - \frac{2 ( 1 - 3 )}{4}}{2}

化简

1 - \frac{2 ( 1 - 3 )}{4} : \frac{4 - 2 ( 1 - 3 )}{4}

1 - \frac{2 ( 1 - 3 )}{4}

将项转换为分式:

1 = \frac{1 \cdot 4}{4}

= \frac{1 \cdot 4}{4} - \frac{2 ( 1 - 3 )}{4}

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 4 - 2 ( 1 - 3 )}{4}

数字相乘:

1 \cdot 4 = 4

= \frac{4 - 2 ( 1 - 3 )}{4}

= \frac{\frac{4 - 2 ( 1 - 3 )}{4}}{2}

使用分式法则:

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

= \frac{4 - 2 ( 1 - 3 )}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{4 - 2 ( 1 - 3 )}{8}

= \frac{4 - 2 ( 1 - 3 )}{8}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{4 - 2 ( 1 - 3 )}{8}

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= \frac{- 2 ( 1 - 3 ) + 4}{2 2}

\frac{4 - 2 ( 1 - 3 )}{2 2}

有理化:

\frac{2 - 2 ( 1 - 3 ) + 4}{4}

\frac{4 - 2 ( 1 - 3 )}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{4 - 2 ( 1 - 3 ) 2}{2 2 2}

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

同类项相加:

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

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

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

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

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2 4 - 2 ( 1 - 3 )}{4}

= \frac{2 - 2 ( 1 - 3 ) + 4}{4}

= \frac{2 4 - 2 ( 1 - 3 )}{4}

分解

4 : 2^{2}

因式分解

4 = 2^{2}

= \frac{2 - 2 ( 1 - 3 ) + 4}{2 ^{2}}

消掉

\frac{2 4 - 2 ( 1 - 3 )}{2 ^{2}} : \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

\frac{2 4 - 2 ( 1 - 3 )}{2 ^{2}}

使用根式运算法则:

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

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

= \frac{2 ^{\frac{1}{2}} - 2 ( 1 - 3 ) + 4}{2 ^{2}}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{4 - 2 ( 1 - 3 )}{2 ^{2 - \frac{1}{2}}}

数字相减:

2 - \frac{1}{2} = \frac{3}{2}

= \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

= \frac{4 - 2 ( 1 - 3 )}{2 ^{\frac{3}{2}}}

2^{\frac{3}{2}} = 22

2^{\frac{3}{2}}

2^{\frac{3}{2}} = 2^{1 + \frac{1}{2}}

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

使用指数法则:

x^{a + b} = x^{a} x^{b}

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

整理后得

= 22

= \frac{4 - 2 ( 1 - 3 )}{2 2}

= \frac{2 ( 1 - 3 ) + 4}{2 2} + \frac{- 2 ( 1 - 3 ) + 4}{2 2} \frac{2 ( 1 - 3 ) + 4}{2 2} - \frac{- 2 ( 1 - 3 ) + 4}{2 2}

化简

\frac{4 + 2 ( 1 - 3 )}{2 2} + \frac{4 - 2 ( 1 - 3 )}{2 2} \frac{4 + 2 ( 1 - 3 )}{2 2} - \frac{4 - 2 ( 1 - 3 )}{2 2}

化简

\frac{2 ( 1 - 3 ) + 4}{2 2} + \frac{- 2 ( 1 - 3 ) + 4}{2 2} : \frac{2 ( 1 - 3 ) + 4 + - 2 ( 1 - 3 ) + 4}{2 2}

\frac{2 ( 1 - 3 ) + 4}{2 2} + \frac{- 2 ( 1 - 3 ) + 4}{2 2}

使用法则

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

= \frac{2 ( 1 - 3 ) + 4 + - 2 ( 1 - 3 ) + 4}{2 2}

= \frac{2 ( 1 - 3 ) + 4 + - 2 ( 1 - 3 ) + 4}{2 2} \frac{2 ( 1 - 3 ) + 4}{2 2} - \frac{- 2 ( 1 - 3 ) + 4}{2 2}

合并分式

\frac{2 ( 1 - 3 ) + 4}{2 2} - \frac{- 2 ( 1 - 3 ) + 4}{2 2} : \frac{4 + 2 ( 1 - 3 ) - 4 - 2 ( 1 - 3 )}{2 2}

使用法则

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

= \frac{2 ( 1 - 3 ) + 4 - - 2 ( 1 - 3 ) + 4}{2 2}

= \frac{2 ( 1 - 3 ) + 4 + - 2 ( 1 - 3 ) + 4}{2 2} \frac{2 ( 1 - 3 ) + 4 - - 2 ( 1 - 3 ) + 4}{2 2}

去除括号:

(a) = a

= \frac{4 + 2 ( 1 - 3 ) + 4 - 2 ( 1 - 3 )}{2 2} \cdot \frac{4 + 2 ( 1 - 3 ) - 4 - 2 ( 1 - 3 )}{2 2}

分式相乘:

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

= \frac{( 4 + 2 ( 1 - 3 ) + 4 - 2 ( 1 - 3 ) ) ( 4 + 2 ( 1 - 3 ) - 4 - 2 ( 1 - 3 ) )}{2 2 \cdot 2 2}

22 \cdot 22 = 8

22 \cdot 22

数字相乘:

2 \cdot 2 = 4

= 422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

4 \cdot 2 = 8

= 8

= \frac{( 2 ( 1 - 3 ) + 4 + - 2 ( 1 - 3 ) + 4 ) ( 2 ( 1 - 3 ) + 4 - - 2 ( 1 - 3 ) + 4 )}{8}

乘开

(4 + 2 (1 - 3) + 4 - 2 (1 - 3)) (4 + 2 (1 - 3) - 4 - 2 (1 - 3)) : 22 - 26

(4 + 2 (1 - 3) + 4 - 2 (1 - 3)) (4 + 2 (1 - 3) - 4 - 2 (1 - 3))

使用平方差公式:

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

a = 4 + 2 (1 - 3), b = 4 - 2 (1 - 3)

= (4 + 2 (1 - 3))^{2} - (4 - 2 (1 - 3))^{2}

化简

(4 + 2 (1 - 3))^{2} - (4 - 2 (1 - 3))^{2} : 22 - 26

(4 + 2 (1 - 3))^{2} - (4 - 2 (1 - 3))^{2}

(4 + 2 (1 - 3))^{2} = 4 + 2 (1 - 3)

(4 + 2 (1 - 3))^{2}

使用根式运算法则:

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

= ((4 + 2 (1 - 3))^{\frac{1}{2}})^{2}

使用指数法则:

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

= (4 + 2 (1 - 3))^{\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

= 4 + 2 (1 - 3)

(4 - 2 (1 - 3))^{2} = 4 - 2 (1 - 3)

(4 - 2 (1 - 3))^{2}

使用根式运算法则:

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

= ((4 - 2 (1 - 3))^{\frac{1}{2}})^{2}

使用指数法则:

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

= (4 - 2 (1 - 3))^{\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

= 4 - 2 (1 - 3)

= 4 + 2 (1 - 3) - (- 2 (1 - 3) + 4)

乘开

2 (1 - 3) : 2 - 6

2 (1 - 3)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = 3

= 2 \cdot 1 - 23

= 1 \cdot 2 - 23

化简

1 \cdot 2 - 23 : 2 - 6

1 \cdot 2 - 23

1 \cdot 2 = 2

1 \cdot 2

乘以:

1 \cdot 2 = 2

= 2

23 = 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= 2 - 6

= 2 - 6

= 4 + 2 - 6 - (4 - 2 (1 - 3))

- (4 - 2 (1 - 3)) : - 4 + 2 (1 - 3)

- (4 - 2 (1 - 3))

打开括号

= - (4) - (- 2 (1 - 3))

使用加减运算法则

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

= - 4 + 2 (1 - 3)

= 4 + 2 - 6 - 4 + 2 (1 - 3)

乘开

2 (1 - 3) : 2 - 6

2 (1 - 3)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = 3

= 2 \cdot 1 - 23

= 1 \cdot 2 - 23

化简

1 \cdot 2 - 23 : 2 - 6

1 \cdot 2 - 23

1 \cdot 2 = 2

1 \cdot 2

乘以:

1 \cdot 2 = 2

= 2

23 = 6

23

使用根式运算法则:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

= 2 - 6

= 2 - 6

= 4 + 2 - 6 - 4 + 2 - 6

化简

4 + 2 - 6 - 4 + 2 - 6 : 22 - 26

4 + 2 - 6 - 4 + 2 - 6

同类项相加:

2 + 2 = 22

= 4 + 22 - 6 - 4 - 6

同类项相加:

- 6 - 6 = - 26

= 4 + 22 - 26 - 4

4 - 4 = 0

= 22 - 26

= 22 - 26

= 22 - 26

= \frac{2 2 - 2 6}{8}

因式分解出通项

2

= \frac{2 ( 2 - 6 )}{8}

约分:

2

= \frac{2 - 6}{4}

= \frac{2 - 6}{4}

= \frac{2 - 6}{4}

= \frac{2 - 6}{4}

调整右侧

cos (\frac{π}{3} + \frac{π}{4})

使用三角恒等式改写

cos (\frac{π}{3} + \frac{π}{4})

使用角和恒等式:

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

= cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4}) = \frac{2 - 6}{4}

cos (\frac{π}{3}) cos (\frac{π}{4}) - sin (\frac{π}{3}) sin (\frac{π}{4})

cos (\frac{π}{3}) cos (\frac{π}{4}) = \frac{2}{4}

cos (\frac{π}{3}) cos (\frac{π}{4})

化简

cos (\frac{π}{3}) : \frac{1}{2}

cos (\frac{π}{3})

使用以下普通恒等式:

cos (\frac{π}{3}) = \frac{1}{2}

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}

= \frac{1}{2}

= \frac{1}{2} cos (\frac{π}{4})

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

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

分式相乘:

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

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

乘以:

1 \cdot 2 = 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

sin (\frac{π}{3}) sin (\frac{π}{4}) = \frac{6}{4}

sin (\frac{π}{3}) sin (\frac{π}{4})

化简

sin (\frac{π}{3}) : \frac{3}{2}

sin (\frac{π}{3})

使用以下普通恒等式:

sin (\frac{π}{3}) = \frac{3}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= \frac{3}{2} sin (\frac{π}{4})

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{3}{2} \cdot \frac{2}{2}

分式相乘:

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

= \frac{3 2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 2}{4}

化简

32 : 6

32

使用根式运算法则:

a b = a \cdot b

32 = 3 \cdot 2

= 3 \cdot 2

数字相乘:

3 \cdot 2 = 6

= 6

= \frac{6}{4}

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

使用法则

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

= \frac{2 - 6}{4}

= \frac{2 - 6}{4}

= \frac{2 - 6}{4}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 csc(θ)=11

p ro v e csc (θ) = 11

证明 cos(pi/2+x)=cos(x)

p ro v e cos (\frac{π}{2} + x) = cos (x)

证明 (csc(6x)+cot(6x))/(csc(6x)-cot(6x))=tan(6x)

p ro v e \frac{csc ( 6 x ) + cot ( 6 x )}{csc ( 6 x ) - cot ( 6 x )} = tan (6 x)

证明 csc^4(x)-cot^4(x)=1+2cot^2(x)

p ro v e csc^{4} (x) - cot^{4} (x) = 1 + 2 cot^{2} (x)

证明 1/(sin(x))sec(x)=tan(x)+cot(x)

p ro v e \frac{1}{sin ( x )} sec (x) = tan (x) + cot (x)