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受欢迎的三角函数 >

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

解答

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

解答

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

+1

度数

x = 68.52929 \dots^{\circ} + 36 0^{\circ} n, x = 291.47070 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

\frac{1}{2}

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

化简

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

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

将项转换为分式:

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

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

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 2 - cos ( x ) \cdot 2 - 1}{2}

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (x)

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

数字相乘:

2 \cdot 1 = 2

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: cos (x) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 (- 2) \cdot 1 = 23

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

数字相加:

4 + 8 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

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

将解分隔开

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

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

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

去除括号:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 + 2 3}{- 4}

使用分式法则:

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

= - \frac{2 + 2 3}{4}

消掉

\frac{2 + 2 3}{4} : \frac{1 + 3}{2}

\frac{2 + 2 3}{4}

分解

2 + 23 : 2 (1 + 3)

2 + 23

改写为

= 2 \cdot 1 + 23

因式分解出通项

2

= 2 (1 + 3)

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

约分:

2

= \frac{1 + 3}{2}

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

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

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

去除括号:

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

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

数字相乘:

2 \cdot 2 = 4

= \frac{2 - 2 3}{- 4}

使用分式法则:

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

2 - 23 = - (23 - 2)

= \frac{2 3 - 2}{4}

分解

23 - 2 : 2 (3 - 1)

23 - 2

改写为

= 23 - 2 \cdot 1

因式分解出通项

2

= 2 (3 - 1)

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

约分:

2

= \frac{3 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 3}{2}, u = \frac{3 - 1}{2}

u = cos (x)

代回

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

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

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

无解

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

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{3 - 1}{2} : x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

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

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

合并所有解

x = arccos (\frac{3 - 1}{2}) + 2 πn, x = 2 π - arccos (\frac{3 - 1}{2}) + 2 πn

以小数形式表示解

x = 1.19606 \dots + 2 πn, x = 2 π - 1.19606 \dots + 2 πn

作图

流行的例子

cos(x)[3sin(x)-2]=0

cos (x) [3 sin (x) - 2] = 0

sin^3(x)+sin(x)-4=0

sin^{3} (x) + sin (x) - 4 = 0

solvefor a,sin^2(a)+cos^2(b)=1

so l v e f or a, sin^{2} (a) + cos^{2} (b) = 1

sin^4(x)=-cos(x)

sin^{4} (x) = - cos (x)

cos^3(x)=4cos^3(x)-3cos(x)

cos^{3} (x) = 4 cos^{3} (x) - 3 cos (x)