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受欢迎的三角函数 >

108cos(0.5a)=sin(90+a)

解答

108 cos (0.5 a) = sin (9 0^{\circ} + a)

解答

a = \frac{1.58005 \dots + 36 0 ^{\circ} n}{0.5}, a = \frac{- 1.58005 \dots + 36 0 ^{\circ} n}{0.5}, a = \frac{\emptyset 10}{5}

+1

弧度

a = 0 + \frac{1.58005 \dots + 2 π}{0.5} n, a = 0 + \frac{- 1.58005 \dots + 2 π}{0.5} n, a = \frac{\emptyset 10}{5}

求解步骤

108 cos (0.5 a) = sin (9 0^{\circ} + a)

使用三角恒等式改写

108 cos (0.5 a) = sin (9 0^{\circ} + a)

使用三角恒等式改写

sin (9 0^{\circ} + a)

使用角和恒等式:

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

= sin (9 0^{\circ}) cos (a) + cos (9 0^{\circ}) sin (a)

化简

sin (9 0^{\circ}) cos (a) + cos (9 0^{\circ}) sin (a) : cos (a)

sin (9 0^{\circ}) cos (a) + cos (9 0^{\circ}) sin (a)

sin (9 0^{\circ}) cos (a) = cos (a)

sin (9 0^{\circ}) cos (a)

化简

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

使用以下普通恒等式:

sin (9 0^{\circ}) = 1

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}

= 1

= 1 \cdot cos (a)

乘以:

1 \cdot cos (a) = cos (a)

= cos (a)

cos (9 0^{\circ}) sin (a) = 0

cos (9 0^{\circ}) sin (a)

化简

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

使用以下普通恒等式:

cos (9 0^{\circ}) = 0

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}

= 0

= 0 \cdot sin (a)

使用法则

0 \cdot a = 0

= 0

= cos (a) + 0

cos (a) + 0 = cos (a)

= cos (a)

= cos (a)

108 cos (0.5 a) = cos (a)

108 cos (0.5 a) = cos (a)

两边减去

cos (a)

108 cos (0.5 a) - cos (a) = 0

令

: u = 0.5 a

108 cos (u) - cos (2 u) = 0

使用三角恒等式改写

- cos (2 u) + 108 cos (u)

使用倍角公式:

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

= - (2 cos^{2} (u) - 1) + 108 cos (u)

- (2 cos^{2} (u) - 1) : - 2 cos^{2} (u) + 1

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

打开括号

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

使用加减运算法则

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

= - 2 cos^{2} (u) + 1

= - 2 cos^{2} (u) + 1 + 108 cos (u)

1 + 108 cos (u) - 2 cos^{2} (u) = 0

用替代法求解

1 + 108 cos (u) - 2 cos^{2} (u) = 0

令

: cos (u) = u

1 + 108 u - 2 u^{2} = 0

1 + 108 u - 2 u^{2} = 0 : u = - \frac{- 54 + 2918}{2}, u = \frac{54 + 2918}{2}

1 + 108 u - 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

- 2 u^{2} + 108 u + 1 = 0

使用求根公式求解

- 2 u^{2} + 108 u + 1 = 0

二次方程求根公式:

若

a = - 2, b = 108, c = 1

u_{1, 2} = \frac{- 108 \pm 10 8 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 108 \pm 10 8 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

10 8^{2} - 4 (- 2) \cdot 1 = 22918

10 8^{2} - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= 10 8^{2} + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 10 8^{2} + 8

10 8^{2} = 11664

= 11664 + 8

数字相加:

11664 + 8 = 11672

= 11672

11672

质因数分解:

2^{3} \cdot 1459

11672

11672

除以

211672 = 5836 \cdot 2

= 2 \cdot 5836

5836

除以

25836 = 2918 \cdot 2

= 2 \cdot 2 \cdot 2918

2918

除以

22918 = 1459 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 1459

2, 1459

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 1459

= 2^{3} \cdot 1459

= 2^{3} \cdot 1459

使用指数法则:

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

= 2^{2} \cdot 2 \cdot 1459

使用根式运算法则:

n ab = n a n b

= 2^{2} 2 \cdot 1459

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 2 \cdot 1459

整理后得

= 22918

u_{1, 2} = \frac{- 108 \pm 2 2918}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 108 + 2 2918}{2 ( - 2 )}, u_{2} = \frac{- 108 - 2 2918}{2 ( - 2 )}

u = \frac{- 108 + 2 2918}{2 ( - 2 )} : - \frac{- 54 + 2918}{2}

\frac{- 108 + 2 2918}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 108 + 2 2918}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 108 + 2 2918}{- 4}

使用分式法则:

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

= - \frac{- 108 + 2 2918}{4}

消掉

\frac{- 108 + 2 2918}{4} : \frac{2918 - 54}{2}

\frac{- 108 + 2 2918}{4}

分解

- 108 + 22918 : 2 (- 54 + 2918)

- 108 + 22918

改写为

= - 2 \cdot 54 + 22918

因式分解出通项

2

= 2 (- 54 + 2918)

= \frac{2 ( - 54 + 2918 )}{4}

约分:

2

= \frac{- 54 + 2918}{2}

= - \frac{2918 - 54}{2}

= - \frac{- 54 + 2918}{2}

u = \frac{- 108 - 2 2918}{2 ( - 2 )} : \frac{54 + 2918}{2}

\frac{- 108 - 2 2918}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 108 - 2 2918}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 108 - 2 2918}{- 4}

使用分式法则:

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

- 108 - 22918 = - (108 + 22918)

= \frac{108 + 2 2918}{4}

分解

108 + 22918 : 2 (54 + 2918)

108 + 22918

改写为

= 2 \cdot 54 + 22918

因式分解出通项

2

= 2 (54 + 2918)

= \frac{2 ( 54 + 2918 )}{4}

约分:

2

= \frac{54 + 2918}{2}

二次方程组的解是:

u = - \frac{- 54 + 2918}{2}, u = \frac{54 + 2918}{2}

u = cos (u)

代回

cos (u) = - \frac{- 54 + 2918}{2}, cos (u) = \frac{54 + 2918}{2}

cos (u) = - \frac{- 54 + 2918}{2}, cos (u) = \frac{54 + 2918}{2}

cos (u) = - \frac{- 54 + 2918}{2} : u = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n, u = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

cos (u) = - \frac{- 54 + 2918}{2}

使用反三角函数性质

cos (u) = - \frac{- 54 + 2918}{2}

cos (u) = - \frac{- 54 + 2918}{2}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 36 0^{\circ} n, x = - arccos (- a) + 36 0^{\circ} n

u = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n, u = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

u = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n, u = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

cos (u) = \frac{54 + 2918}{2} :

无解

cos (u) = \frac{54 + 2918}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

u = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n, u = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

u = 0.5 a

代回

0.5 a = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n : a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

0.5 a = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

两边除以

0.5

0.5 a = arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

两边除以

0.5

\frac{0.5 a}{0.5} = \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

化简

\frac{0.5 a}{0.5} = \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

化简

\frac{0.5 a}{0.5} : a

\frac{0.5 a}{0.5}

约分:

0.5

= a

化简

\frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5} : \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

\frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

使用法则

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

= \frac{arccos ( - \frac{2918 - 54}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

0.5 a = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n : a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

0.5 a = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

两边除以

0.5

0.5 a = - arccos (- \frac{- 54 + 2918}{2}) + 36 0^{\circ} n

两边除以

0.5

\frac{0.5 a}{0.5} = - \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

化简

\frac{0.5 a}{0.5} = - \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

化简

\frac{0.5 a}{0.5} : a

\frac{0.5 a}{0.5}

约分:

0.5

= a

化简

- \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5} : \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

- \frac{arccos ( - \frac{- 54 + 2918}{2} )}{0.5} + \frac{36 0 ^{\circ} n}{0.5}

使用法则

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

= \frac{- arccos ( - \frac{2918 - 54}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}

0.5 a = \emptyset : a = \frac{\emptyset 10}{5}

0.5 a = \emptyset

在两边乘以

10

0.5 a = \emptyset

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

0.5 a \cdot 10 = \emptyset 10

整理后得

5 a = \emptyset 10

5 a = \emptyset 10

两边除以

5

5 a = \emptyset 10

两边除以

5

\frac{5 a}{5} = \frac{\emptyset 10}{5}

化简

a = \frac{\emptyset 10}{5}

a = \frac{\emptyset 10}{5}

a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}, a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}, a = \frac{\emptyset 10}{5}

因为方程对以下值无定义:

\frac{\emptyset 10}{5}

a = \frac{arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}, a = \frac{- arccos ( - \frac{- 54 + 2918}{2} ) + 36 0 ^{\circ} n}{0.5}, a = \frac{\emptyset 10}{5}

以小数形式表示解

a = \frac{1.58005 \dots + 36 0 ^{\circ} n}{0.5}, a = \frac{- 1.58005 \dots + 36 0 ^{\circ} n}{0.5}, a = \frac{\emptyset 10}{5}

作图

流行的例子

cos (a) = \frac{7}{25}

cos (x) = \frac{11}{14}

cot(38)=tan(2x)

cot (3 8^{\circ}) = tan (2 x)

cos (θ) = 0.8126

cos(θ)=-4/5 ,tan(θ)<0

cos (θ) = - \frac{4}{5}, tan (θ) < 0