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受欢迎的三角函数 >

sin^2(x)-cos(2x)=-1/4

解答

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

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

- \frac{1}{4}

sin^{2} (x) - cos (2 x) + \frac{1}{4} = 0

化简

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

sin^{2} (x) - cos (2 x) + \frac{1}{4}

将项转换为分式:

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

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

\frac{4 sin ^{2} ( x ) - 4 cos ( 2 x ) + 1}{4} = 0

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

4 sin^{2} (x) - 4 cos (2 x) + 1 = 0

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

1 - 4 (1 - 2 sin^{2} (x)) + 4 sin^{2} (x) : 12 sin^{2} (x) - 3

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

化简

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

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

数字相乘:

4 \cdot 1 = 4

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

数字相乘:

4 \cdot 2 = 8

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

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

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

化简

1 - 4 + 8 sin^{2} (x) + 4 sin^{2} (x) : 12 sin^{2} (x) - 3

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

同类项相加:

8 sin^{2} (x) + 4 sin^{2} (x) = 12 sin^{2} (x)

= 1 - 4 + 12 sin^{2} (x)

数字相减:

1 - 4 = - 3

= 12 sin^{2} (x) - 3

= 12 sin^{2} (x) - 3

= 12 sin^{2} (x) - 3

- 3 + 12 sin^{2} (x) = 0

用替代法求解

- 3 + 12 sin^{2} (x) = 0

令

: sin (x) = u

- 3 + 12 u^{2} = 0

- 3 + 12 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

- 3 + 12 u^{2} = 0

将

3

到右边

- 3 + 12 u^{2} = 0

两边加上

3

- 3 + 12 u^{2} + 3 = 0 + 3

化简

12 u^{2} = 3

12 u^{2} = 3

两边除以

12

12 u^{2} = 3

两边除以

12

\frac{12 u ^{2}}{12} = \frac{3}{12}

化简

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{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{1}{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{1}{2}

使用法则

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

化简

\frac{1}{4} : \frac{1}{2}

\frac{1}{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{1}{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{1}{2}

= - \frac{1}{2}

使用法则

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

作图

流行的例子

2=cos^2(x)+sin^2(x)

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

-5pisin((pi(t))/2)=7.071

- 5 π sin (\frac{π ( t )}{2}) = 7.071

9.8*sin(x)-19.6*cos(x)=4.7

9.8 \cdot sin (x) - 19.6 \cdot cos (x) = 4.7

cos(x)=-sqrt(3)sin(x)

cos (x) = - 3 sin (x)

sin(α)= 3/5 ,0<a< pi/2

sin (α) = \frac{3}{5}, 0 < a < \frac{π}{2}