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受欢迎的三角函数 >

sec((15pi)/8)

解答

sec (\frac{15 π}{8})

解答

2 2 + 2 - 2 2 + 2

+1

十进制

1.08239 \dots

求解步骤

sec (\frac{15 π}{8})

使用三角恒等式改写:

\frac{1}{cos ( \frac{15 π}{8} )}

sec (\frac{15 π}{8})

使用基本三角恒等式:

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

= \frac{1}{cos ( \frac{15 π}{8} )}

= \frac{1}{cos ( \frac{15 π}{8} )}

使用三角恒等式改写:

cos (\frac{15 π}{8}) = \frac{2 + 2}{2}

cos (\frac{15 π}{8})

使用三角恒等式改写:

\frac{1 + cos ( \frac{7 π}{4} )}{2}

cos (\frac{15 π}{8})

将

cos (\frac{15 π}{8})

写为

cos (\frac{\frac{15 π}{4}}{2})

= cos (\frac{\frac{15 π}{4}}{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{15 π}{4} )}{2}

cos (\frac{15 π}{4}) = cos (\frac{7 π}{4})

cos (\frac{15 π}{4})

将

\frac{15 π}{4}

改写为

2 π + \frac{7 π}{4}

= cos (2 π + \frac{7 π}{4})

使用周期

cos

:

cos (x + 2 π) = cos (x)

cos (2 π + \frac{7 π}{4}) = cos (\frac{7 π}{4})

= cos (\frac{7 π}{4})

= \frac{1 + cos ( \frac{7 π}{4} )}{2}

= \frac{1 + cos ( \frac{7 π}{4} )}{2}

使用三角恒等式改写:

cos (\frac{7 π}{4}) = \frac{2}{2}

cos (\frac{7 π}{4})

使用三角恒等式改写:

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

cos (\frac{7 π}{4})

将

cos (\frac{7 π}{4})

写为

cos (π + \frac{3 π}{4})

= cos (π + \frac{3 π}{4})

使用角和恒等式:

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

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

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

使用以下普通恒等式:

cos (π) = (- 1)

cos (π)

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}

= (- 1)

使用以下普通恒等式:

cos (\frac{3 π}{4}) = - \frac{2}{2}

cos (\frac{3 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 (π) = 0

sin (π)

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}

= 0

使用以下普通恒等式:

sin (\frac{3 π}{4}) = \frac{2}{2}

sin (\frac{3 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}

= (- 1) (- \frac{2}{2}) - 0 \cdot \frac{2}{2}

化简

= \frac{2}{2}

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

化简

\frac{1 + \frac{2}{2}}{2} : \frac{2 + 2}{2}

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

\frac{1 + \frac{2}{2}}{2} = \frac{2 + 2}{4}

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

化简

1 + \frac{2}{2} : \frac{2 + 2}{2}

1 + \frac{2}{2}

将项转换为分式:

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

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

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 2 + 2}{2}

数字相乘:

1 \cdot 2 = 2

= \frac{2 + 2}{2}

= \frac{\frac{2 + 2}{2}}{2}

使用分式法则:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 2}{4}

= \frac{2 + 2}{4}

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{2 + 2}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 2}{2}

= \frac{2 + 2}{2}

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

化简

\frac{1}{\frac{2 + 2}{2}} : 2 2 + 2 - 2 2 + 2

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

使用分式法则:

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

= \frac{2}{2 + 2}

\frac{2}{2 + 2}

有理化:

2 2 + 2 - 2 2 + 2

\frac{2}{2 + 2}

乘以共轭根式

\frac{2 + 2}{2 + 2}

= \frac{2 2 + 2}{2 + 2 2 + 2}

2 + 2 2 + 2 = 2 + 2

2 + 2 2 + 2

使用根式运算法则:

a a = a

2 + 2 2 + 2 = 2 + 2

= 2 + 2

= \frac{2 2 + 2}{2 + 2}

分解

2 + 2 : 2 (2 + 1)

2 + 2

2 = 22

= 22 + 2

因式分解出通项

2

= 2 (2 + 1)

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

消掉

\frac{2 2 + 2}{2 ( 2 + 1 )} : \frac{2 2 + 2}{2 + 1}

\frac{2 2 + 2}{2 ( 2 + 1 )}

使用根式运算法则:

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

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

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

使用指数法则:

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

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

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

数字相减:

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

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

使用根式运算法则:

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

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

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

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

乘以共轭根式

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

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

2 2 + 2 (2 - 1) = 2 2 + 2 - 2 2 + 2

2 2 + 2 (2 - 1)

= 2 (2 - 1) 2 + 2

使用分配律:

a (b - c) = ab - a c

a = 2 2 + 2, b = 2, c = 1

= 2 2 + 2 2 - 2 2 + 2 \cdot 1

= 22 2 + 2 - 1 \cdot 2 2 + 2

化简

22 2 + 2 - 1 \cdot 2 2 + 2 : 2 2 + 2 - 2 2 + 2

22 2 + 2 - 1 \cdot 2 2 + 2

使用根式运算法则:

a a = a

22 = 2

= 2 2 + 2 - 1 \cdot 2 2 + 2

乘以:

1 \cdot 2 = 2

= 2 2 + 2 - 2 2 + 2

= 2 2 + 2 - 2 2 + 2

(2 + 1) (2 - 1) = 1

(2 + 1) (2 - 1)

使用平方差公式:

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

a = 2, b = 1

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

化简

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

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

使用法则

1^{a} = 1

1^{2} = 1

= (2)^{2} - 1

(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

= 2 - 1

数字相减:

2 - 1 = 1

= 1

= 1

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

使用法则

\frac{a}{1} = a

= 2 2 + 2 - 2 2 + 2

= 2 2 + 2 - 2 2 + 2

= 2 2 + 2 - 2 2 + 2

流行的例子

tan (- 22. 5^{\circ})

- 4 sin (0)

arcsin(cos(-(5pi)/6))

arcsin (cos (- \frac{5 π}{6}))

tan(arctan(pi))

tan (arctan (π))

cos(pi/6)+isin(pi/6)

cos (\frac{π}{6}) + i sin (\frac{π}{6})