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受欢迎的 >

sec^2(pi/5)

解答

sec^{2} (\frac{π}{5})

解答

6 - 25

+1

十进制表示法

1.52786 \dots

求解步骤

sec^{2} (\frac{π}{5})

使用三角恒等式改写:

sec (\frac{π}{5}) = 5 - 1

sec (\frac{π}{5})

使用三角恒等式改写:

\frac{1}{cos ( \frac{π}{5} )}

sec (\frac{π}{5})

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{cos ( \frac{π}{5} )}

= \frac{1}{cos ( \frac{π}{5} )}

使用三角恒等式改写:

cos (\frac{π}{5}) = \frac{5 + 1}{4}

cos (\frac{π}{5})

显示:

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

使用以下积化和差公式:

2 sin (x) cos (y) = sin (x + y) - sin (x - y)

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

\frac{1}{2} = sin (\frac{3 π}{10}) - sin (\frac{π}{10})

sin (\frac{3 π}{10}) = cos (\frac{π}{2} - \frac{3 π}{10})

\frac{1}{2} = cos (\frac{π}{2} - \frac{3 π}{10}) - sin (\frac{π}{10})

\frac{1}{2} = cos (\frac{π}{5}) - sin (\frac{π}{10})

显示:

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

使用因式分解法则:

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

a = cos (\frac{π}{5}) + sin (\frac{π}{10})

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = ((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) ((cos (\frac{π}{5}) + sin (\frac{π}{10})) - (cos (\frac{π}{5}) - sin (\frac{π}{10})))

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 2 (2 cos (\frac{π}{5}) sin (\frac{π}{10}))

显示:

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

sin (\frac{2 π}{5}) = 2 sin (\frac{π}{5}) cos (\frac{π}{5})

sin (\frac{2 π}{5}) sin (\frac{π}{5}) = 4 sin (\frac{π}{5}) sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

sin (\frac{π}{5})

sin (\frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

利用以下特性:

sin (x) = cos (\frac{π}{2} - x)

sin (\frac{2 π}{5}) = cos (\frac{π}{2} - \frac{2 π}{5})

cos (\frac{π}{2} - \frac{2 π}{5}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

cos (\frac{π}{10}) = 4 sin (\frac{π}{10}) cos (\frac{π}{5}) cos (\frac{π}{10})

两边除以

cos (\frac{π}{10})

1 = 4 sin (\frac{π}{10}) cos (\frac{π}{5})

两边除以

2

\frac{1}{2} = 2 sin (\frac{π}{10}) cos (\frac{π}{5})

代入

2 cos (\frac{π}{5}) sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (cos (\frac{π}{5}) - sin (\frac{π}{10}))^{2} = 1

代入

cos (\frac{π}{5}) - sin (\frac{π}{10}) = \frac{1}{2}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - (\frac{1}{2})^{2} = 1

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} = 1

两边加上

\frac{1}{4}

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} - \frac{1}{4} + \frac{1}{4} = 1 + \frac{1}{4}

整理后得

(cos (\frac{π}{5}) + sin (\frac{π}{10}))^{2} = \frac{5}{4}

在两侧开平方

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \pm \frac{5}{4}

cos (\frac{π}{5})

不能为负

sin (\frac{π}{10})

不能为负

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{4}

以下方程式相加

cos (\frac{π}{5}) + sin (\frac{π}{10}) = \frac{5}{2}

((cos (\frac{π}{5}) + sin (\frac{π}{10})) + (cos (\frac{π}{5}) - sin (\frac{π}{10}))) = (\frac{5}{2} + \frac{1}{2})

整理后得

cos (\frac{π}{5}) = \frac{5 + 1}{4}

= \frac{5 + 1}{4}

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

化简

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

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

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{4}{5 + 1}

\frac{4}{5 + 1}

有理化:

5 - 1

\frac{4}{5 + 1}

乘以共轭根式

\frac{5 - 1}{5 - 1}

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

(5 + 1) (5 - 1) = 4

(5 + 1) (5 - 1)

使用平方差公式:

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

a = 5, b = 1

= (5)^{2} - 1^{2}

化简

(5)^{2} - 1^{2} : 4

(5)^{2} - 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (5)^{2} - 1

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

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

= (5^{\frac{1}{2}})^{2}

使用指数法则:

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

= 5^{\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

= 5

= 5 - 1

数字相减:

5 - 1 = 4

= 4

= 4

= \frac{4 ( 5 - 1 )}{4}

数字相除:

\frac{4}{4} = 1

= 5 - 1

= 5 - 1

= 5 - 1

= (5 - 1)^{2}

化简

(5 - 1)^{2} : 6 - 25

(5 - 1)^{2}

使用完全平方公式:

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

a = 5, b = 1

= (5)^{2} - 25 \cdot 1 + 1^{2}

化简

(5)^{2} - 25 \cdot 1 + 1^{2} : 6 - 25

(5)^{2} - 25 \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (5)^{2} - 2 \cdot 1 \cdot 5 + 1

(5)^{2} = 5

(5)^{2}

使用根式运算法则:

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

= (5^{\frac{1}{2}})^{2}

使用指数法则:

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

= 5^{\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

= 5

25 \cdot 1 = 25

25 \cdot 1

数字相乘:

2 \cdot 1 = 2

= 25

= 5 - 25 + 1

数字相加:

5 + 1 = 6

= 6 - 25

= 6 - 25

= 6 - 25

流行的例子

cot (45 0^{\circ})

cos (\frac{1}{6} π)

sech (1)

arctan (- \frac{6}{5})

\frac{12}{sin ( 5 3 ^{\circ} )}