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受欢迎的三角函数 >

sin(x)+pi/2 =cos(x)

解答

sin (x) + \frac{π}{2} = cos (x)

解答

x \in R 无解

求解步骤

sin (x) + \frac{π}{2} = cos (x)

两边进行平方

(sin (x) + \frac{π}{2})^{2} = cos^{2} (x)

两边减去

cos^{2} (x)

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x) = 0

化简

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x) : \frac{4 sin ^{2} ( x ) + 4 π sin ( x ) + π ^{2} - 4 cos ^{2} ( x )}{4}

sin^{2} (x) + π sin (x) + \frac{π ^{2}}{4} - cos^{2} (x)

将项转换为分式:

sin^{2} (x) = \frac{sin ^{2} ( x ) 4}{4}, π sin (x) = \frac{π sin ( x ) 4}{4}, cos^{2} (x) = \frac{cos ^{2} ( x ) 4}{4}

= \frac{sin ^{2} ( x ) \cdot 4}{4} + \frac{π sin ( x ) \cdot 4}{4} + \frac{π ^{2}}{4} - \frac{cos ^{2} ( x ) \cdot 4}{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{sin ^{2} ( x ) \cdot 4 + π sin ( x ) \cdot 4 + π ^{2} - cos ^{2} ( x ) \cdot 4}{4}

\frac{4 sin ^{2} ( x ) + 4 π sin ( x ) + π ^{2} - 4 cos ^{2} ( x )}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (x) + 4 π sin (x) + π^{2} - 4 cos^{2} (x) = 0

使用三角恒等式改写

π^{2} - 4 cos^{2} (x) + 4 sin^{2} (x) + 4 sin (x) π

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

= π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π

化简

π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π : π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 sin (x) π

= π^{2} - 4 (1 - sin^{2} (x)) + 4 sin^{2} (x) + 4 π sin (x)

乘开

- 4 (1 - sin^{2} (x)) : - 4 + 4 sin^{2} (x)

- 4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = sin^{2} (x)

= - 4 \cdot 1 - (- 4) sin^{2} (x)

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 sin^{2} (x)

= π^{2} - 4 + 4 sin^{2} (x) + 4 sin^{2} (x) + 4 sin (x) π

同类项相加:

4 sin^{2} (x) + 4 sin^{2} (x) = 8 sin^{2} (x)

= π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

= π^{2} - 4 + 8 sin^{2} (x) + 4 π sin (x)

- 4 + π^{2} + 8 sin^{2} (x) + 4 sin (x) π = 0

用替代法求解

- 4 + π^{2} + 8 sin^{2} (x) + 4 sin (x) π = 0

令

: sin (x) = u

- 4 + π^{2} + 8 u^{2} + 4 u π = 0

- 4 + π^{2} + 8 u^{2} + 4 u π = 0 : u = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, u = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

- 4 + π^{2} + 8 u^{2} + 4 u π = 0

改写成标准形式

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

8 u^{2} + 4 π u - 4 + π^{2} = 0

使用求根公式求解

8 u^{2} + 4 π u - 4 + π^{2} = 0

二次方程求根公式:

若

a = 8, b = 4 π, c = - 4 + π^{2}

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

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

化简

(4 π)^{2} - 4 \cdot 8 (- 4 + π^{2}) : i 16 π^{2} - 128

(4 π)^{2} - 4 \cdot 8 (- 4 + π^{2})

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} π^{2} - 4 \cdot 8 (π^{2} - 4)

数字相乘:

4 \cdot 8 = 32

= 4^{2} π^{2} - 32 (π^{2} - 4)

使用虚数运算法则:

- a = i a

= i 32 (π^{2} - 4) - 4^{2} π^{2}

- 4^{2} π^{2} + 32 (- 4 + π^{2}) = 16 π^{2} - 128

- 4^{2} π^{2} + 32 (- 4 + π^{2})

4^{2} = 16

= - 16 π^{2} + 32 (π^{2} - 4)

乘开

- 16 π^{2} + 32 (- 4 + π^{2}) : 16 π^{2} - 128

- 16 π^{2} + 32 (- 4 + π^{2})

乘开

32 (- 4 + π^{2}) : - 128 + 32 π^{2}

32 (- 4 + π^{2})

使用分配律:

a (b + c) = ab + a c

a = 32, b = - 4, c = π^{2}

= 32 (- 4) + 32 π^{2}

使用加减运算法则

+ (- a) = - a

= - 32 \cdot 4 + 32 π^{2}

数字相乘:

32 \cdot 4 = 128

= - 128 + 32 π^{2}

= - 16 π^{2} - 128 + 32 π^{2}

化简

- 16 π^{2} - 128 + 32 π^{2} : 16 π^{2} - 128

- 16 π^{2} - 128 + 32 π^{2}

对同类项分组

= - 16 π^{2} + 32 π^{2} - 128

同类项相加:

- 16 π^{2} + 32 π^{2} = 16 π^{2}

= 16 π^{2} - 128

= 16 π^{2} - 128

= 16 π^{2} - 128

= i 16 π^{2} - 128

u_{1, 2} = \frac{- 4 π \pm i 16 π ^{2} - 128}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8}, u_{2} = \frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8}

u = \frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8} : - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}

\frac{- 4 π + i 16 π ^{2} - 128}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4 π + i 16 π ^{2} - 128}{16}

将

\frac{- 4 π + i 16 π ^{2} - 128}{16}

改写成标准复数形式:

- \frac{π}{4} + \frac{16 π ^{2} - 128}{16} i

\frac{- 4 π + i 16 π ^{2} - 128}{16}

使用分式法则:

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

\frac{- 4 π + i 16 π ^{2} - 128}{16} = - \frac{4 π}{16} + \frac{i 16 π ^{2} - 128}{16}

= - \frac{4 π}{16} + \frac{i 16 π ^{2} - 128}{16}

消掉

\frac{4 π}{16} : \frac{π}{4}

\frac{4 π}{16}

约分:

4

= \frac{π}{4}

= - \frac{π}{4} + \frac{i 16 π ^{2} - 128}{16}

= - \frac{π}{4} + \frac{16 π ^{2} - 128}{16} i

u = \frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8} : - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

\frac{- 4 π - i 16 π ^{2} - 128}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 4 π - i 16 π ^{2} - 128}{16}

将

\frac{- 4 π - i 16 π ^{2} - 128}{16}

改写成标准复数形式:

- \frac{π}{4} - \frac{16 π ^{2} - 128}{16} i

\frac{- 4 π - i 16 π ^{2} - 128}{16}

使用分式法则:

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

\frac{- 4 π - i 16 π ^{2} - 128}{16} = - \frac{4 π}{16} - \frac{i 16 π ^{2} - 128}{16}

= - \frac{4 π}{16} - \frac{i 16 π ^{2} - 128}{16}

消掉

\frac{4 π}{16} : \frac{π}{4}

\frac{4 π}{16}

约分:

4

= \frac{π}{4}

= - \frac{π}{4} - \frac{i 16 π ^{2} - 128}{16}

= - \frac{π}{4} - \frac{16 π ^{2} - 128}{16} i

二次方程组的解是:

u = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, u = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

u = sin (x)

代回

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}, sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16} :

无解

sin (x) = - \frac{π}{4} + i \frac{- 128 + 16 π ^{2}}{16}

无解

sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16} :

无解

sin (x) = - \frac{π}{4} - i \frac{- 128 + 16 π ^{2}}{16}

无解

合并所有解

无解

将解代入原方程进行验证

将它们代入

sin (x) + \frac{π}{2} = cos (x)

检验解是否符合

去除与方程不符的解。

x \in R 无解

作图

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