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受欢迎的三角函数 >

sin(x)-cos(x)-1/(sqrt(2))=0

解答

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

解答

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

+1

度数

x = 7 5^{\circ} + 36 0^{\circ} n, x = 19 5^{\circ} + 36 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

证明恒等式：

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

sin (x) - cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

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

使用角和恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= 2 sin (x - \frac{π}{4})

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

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

将

\frac{1}{2}

到右边

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

两边加上

\frac{1}{2}

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

化简

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

化简

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

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

同类项相加:

- \frac{1}{2} + \frac{1}{2} = 0

= 2 sin (x - \frac{π}{4})

化简

0 + \frac{1}{2} : \frac{2}{2}

0 + \frac{1}{2}

0 + \frac{1}{2} = \frac{1}{2}

= \frac{1}{2}

\frac{1}{2}

有理化:

\frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

2 sin (x - \frac{π}{4}) = \frac{2}{2}

2 sin (x - \frac{π}{4}) = \frac{2}{2}

2 sin (x - \frac{π}{4}) = \frac{2}{2}

两边除以

2

2 sin (x - \frac{π}{4}) = \frac{2}{2}

两边除以

2

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{\frac{2}{2}}{2}

化简

\frac{2 sin ( x - \frac{π}{4} )}{2} = \frac{\frac{2}{2}}{2}

化简

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

\frac{2 sin ( x - \frac{π}{4} )}{2}

约分:

2

= sin (x - \frac{π}{4})

化简

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

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

使用分式法则:

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

= \frac{2}{2 2}

\frac{2}{2 2}

有理化:

\frac{1}{2}

\frac{2}{2 2}

乘以共轭根式

\frac{2}{2}

= \frac{2 2}{2 2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

222 = 4

222

使用指数法则:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

= 2^{1 + \frac{1}{2} + \frac{1}{2}}

同类项相加:

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

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

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

2 \cdot \frac{1}{2}

分式相乘:

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

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

约分:

2

= 1

= 2^{1 + 1}

数字相加:

1 + 1 = 2

= 2^{2}

2^{2} = 4

= 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

= \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \frac{1}{2}

sin (x - \frac{π}{4}) = \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}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

x - \frac{π}{4} = \frac{π}{6} + 2 πn, x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

解

x - \frac{π}{4} = \frac{π}{6} + 2 πn : x = 2 πn + \frac{5 π}{12}

x - \frac{π}{4} = \frac{π}{6} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{π}{6} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{5 π}{12}

\frac{π}{6} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{6} + \frac{π}{4}

6, 4

的最小公倍数:

12

6, 4

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

将每个因子乘以它在

6

或

4

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{6} :

将分母和分子乘以

2

\frac{π}{6} = \frac{π 2}{6 \cdot 2} = \frac{π 2}{12}

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

= \frac{π 2}{12} + \frac{π 3}{12}

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{π 2 + π 3}{12}

同类项相加:

2 π + 3 π = 5 π

= 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

x = 2 πn + \frac{5 π}{12}

解

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn : x = 2 πn + \frac{13 π}{12}

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{5 π}{6} + 2 πn

两边加上

\frac{π}{4}

x - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} = \frac{5 π}{6} + 2 πn + \frac{π}{4}

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

\frac{5 π}{6} + 2 πn + \frac{π}{4} : 2 πn + \frac{13 π}{12}

\frac{5 π}{6} + 2 πn + \frac{π}{4}

对同类项分组

= 2 πn + \frac{π}{4} + \frac{5 π}{6}

4, 6

的最小公倍数:

12

4, 6

最小公倍数 (LCM)

4

质因数分解:

2 \cdot 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

4

或

6

中出现的最多次数

= 2 \cdot 2 \cdot 3

数字相乘:

2 \cdot 2 \cdot 3 = 12

= 12

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

12

对于

\frac{π}{4} :

将分母和分子乘以

3

\frac{π}{4} = \frac{π 3}{4 \cdot 3} = \frac{π 3}{12}

对于

\frac{5 π}{6} :

将分母和分子乘以

2

\frac{5 π}{6} = \frac{5 π 2}{6 \cdot 2} = \frac{10 π}{12}

= \frac{π 3}{12} + \frac{10 π}{12}

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{π 3 + 10 π}{12}

同类项相加:

3 π + 10 π = 13 π

= 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{13 π}{12}

x = 2 πn + \frac{5 π}{12}, x = 2 πn + \frac{13 π}{12}

作图

流行的例子

cos^2(θ)= 49/64

cos^{2} (θ) = \frac{49}{64}

sec^2(θ)=1-tan(θ)

sec^{2} (θ) = 1 - tan (θ)

cos^2(x)=-0.267

cos^{2} (x) = - 0.267

sin(x-4)=cos(9x+4)

sin (x - 4) = cos (9 x + 4)

solvefor x,0=-2sin(x)-4cos(2x)

so l v e f or x, 0 = - 2 sin (x) - 4 cos (2 x)