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受欢迎的三角函数 >

sin((13pi)/8)

解答

sin (\frac{13 π}{8})

解答

- \frac{2 + 2}{2}

+1

十进制

- 0.92387 \dots

求解步骤

sin (\frac{13 π}{8})

使用三角恒等式改写:

- \frac{1 - cos ( \frac{5 π}{4} )}{2}

sin (\frac{13 π}{8})

将

sin (\frac{13 π}{8})

写为

sin (\frac{\frac{13 π}{4}}{2})

= sin (\frac{\frac{13 π}{4}}{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{13 π}{4} )}{2}

cos (\frac{13 π}{4}) = cos (\frac{5 π}{4})

cos (\frac{13 π}{4})

将

\frac{13 π}{4}

改写为

2 π + \frac{5 π}{4}

= cos (2 π + \frac{5 π}{4})

使用周期

cos

:

cos (x + 2 π) = cos (x)

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

= cos (\frac{5 π}{4})

= - \frac{1 - cos ( \frac{5 π}{4} )}{2}

= - \frac{1 - cos ( \frac{5 π}{4} )}{2}

使用三角恒等式改写:

cos (\frac{5 π}{4}) = - \frac{2}{2}

cos (\frac{5 π}{4})

使用三角恒等式改写:

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

cos (\frac{5 π}{4})

将

cos (\frac{5 π}{4})

写为

cos (π + \frac{π}{4})

= cos (π + \frac{π}{4})

使用角和恒等式:

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

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

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

= (- 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}

使用法则

- (- a) = a

= - \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}

化简

\frac{2 + 2}{4} : \frac{2 + 2}{2}

\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{2 + 2}{2}

流行的例子

sec (\frac{15 π}{8})

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

- 4 sin (0)

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

arcsin (cos (- \frac{5 π}{6}))

tan(arctan(pi))

tan (arctan (π))