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受欢迎的三角函数 >

sin(97.5)

解答

sin (97. 5^{\circ})

解答

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

+1

十进制

0.99144 \dots

求解步骤

sin (97. 5^{\circ})

使用三角恒等式改写:

\frac{1 - cos ( 19 5 ^{\circ} )}{2}

sin (97. 5^{\circ})

将

sin (97. 5^{\circ})

写为

sin (\frac{19 5 ^{\circ}}{2})

= sin (\frac{19 5 ^{\circ}}{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, 9 0^{\circ}] [9 0^{\circ}, 18 0^{\circ}] [18 0^{\circ}, 27 0^{\circ}] [27 0^{\circ}, 36 0^{\circ}] 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 ( 19 5 ^{\circ} )}{2}

= \frac{1 - cos ( 19 5 ^{\circ} )}{2}

使用三角恒等式改写:

cos (19 5^{\circ}) = \frac{2 ( - 3 - 1 )}{4}

cos (19 5^{\circ})

使用三角恒等式改写:

cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

cos (19 5^{\circ})

将

cos (19 5^{\circ})

写为

cos (15 0^{\circ} + 4 5^{\circ})

= cos (15 0^{\circ} + 4 5^{\circ})

使用角和恒等式:

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

= cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

= cos (15 0^{\circ}) cos (4 5^{\circ}) - sin (15 0^{\circ}) sin (4 5^{\circ})

使用以下普通恒等式:

cos (15 0^{\circ}) = - \frac{3}{2}

cos (15 0^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{3}{2}

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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 (15 0^{\circ}) = \frac{1}{2}

sin (15 0^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (4 5^{\circ})

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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}) \frac{2}{2} - \frac{1}{2} \cdot \frac{2}{2}

化简

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

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

去除括号:

(- a) = - a

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

因式分解出通项

\frac{2}{2}

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

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

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

使用法则

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

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

数字相乘:

2 \cdot 2 = 4

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

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

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

化简

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

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

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

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

化简

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

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

将项转换为分式:

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

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

数字相乘:

4 \cdot 2 = 8

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

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

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

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

流行的例子

sin (\frac{23 π}{2})

coth (3)

(sin(30))/(sin(20.9))

\frac{sin ( 3 0 ^{\circ} )}{sin ( 20. 9 ^{\circ} )}

1/2 arctan(1/4)

\frac{1}{2} arctan (\frac{1}{4})

arcsin(0.49969)

arcsin (0.49969)