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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

- cos^{2} (x) - 1

- sin (x) + cos^{2} (x) + 1 = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

2 - sin (x) - sin^{2} (x) = 0

用替代法求解

2 - sin (x) - sin^{2} (x) = 0

令

: sin (x) = u

2 - u - u^{2} = 0

2 - u - u^{2} = 0 : u = - 2, u = 1

2 - u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 1 )}

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

1 + 3 = 4

= \frac{4}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{- 2}

使用分式法则:

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

= - \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= - 2

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

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

去除括号:

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

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{- 2}

使用分式法则:

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

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - 2, u = 1

u = sin (x)

代回

sin (x) = - 2, sin (x) = 1

sin (x) = - 2, sin (x) = 1

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

的通解

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}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn

作图

流行的例子

2 sin (θ) - 1 = 0

sin((19pi)/(12))

sin (\frac{19 π}{12})

sin (36 0^{\circ})

sin (θ) = - \frac{1}{2}

sin (15 0^{\circ})