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受欢迎的三角函数 >

tan(45-x)+tan(x)=1

解答

tan (4 5^{\circ} - x) + tan (x) = 1

解答

x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

+1

弧度

x = 0 + 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

求解步骤

tan (4 5^{\circ} - x) + tan (x) = 1

使用三角恒等式改写

tan (4 5^{\circ} - x) + tan (x) = 1

使用三角恒等式改写

tan (4 5^{\circ} - x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( 4 5 ^{\circ} - x )}{cos ( 4 5 ^{\circ} - x )}

使用角差恒等式:

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

= \frac{sin ( 4 5 ^{\circ} ) cos ( x ) - cos ( 4 5 ^{\circ} ) sin ( x )}{cos ( 4 5 ^{\circ} - x )}

使用角差恒等式:

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

= \frac{sin ( 4 5 ^{\circ} ) cos ( x ) - cos ( 4 5 ^{\circ} ) sin ( x )}{cos ( 4 5 ^{\circ} ) cos ( x ) + sin ( 4 5 ^{\circ} ) sin ( x )}

化简

\frac{sin ( 4 5 ^{\circ} ) cos ( x ) - cos ( 4 5 ^{\circ} ) sin ( x )}{cos ( 4 5 ^{\circ} ) cos ( x ) + sin ( 4 5 ^{\circ} ) sin ( x )} : \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

\frac{sin ( 4 5 ^{\circ} ) cos ( x ) - cos ( 4 5 ^{\circ} ) sin ( x )}{cos ( 4 5 ^{\circ} ) cos ( x ) + sin ( 4 5 ^{\circ} ) sin ( x )}

sin (4 5^{\circ}) cos (x) - cos (4 5^{\circ}) sin (x) = \frac{2}{2} cos (x) - \frac{2}{2} sin (x)

sin (4 5^{\circ}) cos (x) - cos (4 5^{\circ}) sin (x)

化简

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} cos (x) - cos (4 5^{\circ}) sin (x)

化简

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

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

= \frac{\frac{2}{2} cos ( x ) - \frac{2}{2} sin ( x )}{cos ( 4 5 ^{\circ} ) cos ( x ) + sin ( 4 5 ^{\circ} ) sin ( x )}

cos (4 5^{\circ}) cos (x) + sin (4 5^{\circ}) sin (x) = \frac{2}{2} cos (x) + \frac{2}{2} sin (x)

cos (4 5^{\circ}) cos (x) + sin (4 5^{\circ}) sin (x)

化简

cos (4 5^{\circ}) : \frac{2}{2}

cos (4 5^{\circ})

使用以下普通恒等式:

cos (4 5^{\circ}) = \frac{2}{2}

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (x) + sin (4 5^{\circ}) sin (x)

化简

sin (4 5^{\circ}) : \frac{2}{2}

sin (4 5^{\circ})

使用以下普通恒等式:

sin (4 5^{\circ}) = \frac{2}{2}

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

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

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

乘

\frac{2}{2} cos (x) : \frac{2 cos ( x )}{2}

\frac{2}{2} cos (x)

分式相乘:

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

= \frac{2 cos ( x )}{2}

= \frac{\frac{2}{2} cos ( x ) - \frac{2}{2} sin ( x )}{\frac{2 c o s ( x )}{2} + \frac{2}{2} sin ( x )}

乘

\frac{2}{2} sin (x) : \frac{2 sin ( x )}{2}

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

分式相乘:

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

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

= \frac{\frac{2}{2} cos ( x ) - \frac{2}{2} sin ( x )}{\frac{2 c o s ( x )}{2} + \frac{2 s i n ( x )}{2}}

乘

\frac{2}{2} cos (x) : \frac{2 cos ( x )}{2}

\frac{2}{2} cos (x)

分式相乘:

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

= \frac{2 cos ( x )}{2}

= \frac{\frac{2 c o s ( x )}{2} - \frac{2}{2} sin ( x )}{\frac{2 c o s ( x )}{2} + \frac{2 s i n ( x )}{2}}

乘

\frac{2}{2} sin (x) : \frac{2 sin ( x )}{2}

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

分式相乘:

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

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

= \frac{\frac{2 c o s ( x )}{2} - \frac{2 s i n ( x )}{2}}{\frac{2 c o s ( x )}{2} + \frac{2 s i n ( x )}{2}}

合并分式

\frac{2 cos ( x )}{2} + \frac{2 sin ( x )}{2} : \frac{2 cos ( x ) + 2 sin ( x )}{2}

使用法则

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

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

= \frac{\frac{2 c o s ( x )}{2} - \frac{2 s i n ( x )}{2}}{\frac{2 c o s ( x ) + 2 s i n ( x )}{2}}

合并分式

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

使用法则

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

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

= \frac{\frac{2 c o s ( x ) - 2 s i n ( x )}{2}}{\frac{2 c o s ( x ) + 2 s i n ( x )}{2}}

分式相除:

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

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

约分:

2

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

因式分解出通项

2

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

因式分解出通项

2

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

约分:

2

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )}

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

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

两边减去

1

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

化简

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

\frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )} + tan (x) - 1

将项转换为分式:

tan (x) = \frac{tan ( x ) ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )}, 1 = \frac{1 ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )}

= \frac{cos ( x ) - sin ( x )}{cos ( x ) + sin ( x )} + \frac{tan ( x ) ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )} - \frac{1 \cdot ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )}

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{cos ( x ) - sin ( x ) + tan ( x ) ( cos ( x ) + sin ( x ) ) - 1 \cdot ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )}

乘以:

1 \cdot (cos (x) + sin (x)) = (cos (x) + sin (x))

= \frac{cos ( x ) - sin ( x ) + tan ( x ) ( cos ( x ) + sin ( x ) ) - ( cos ( x ) + sin ( x ) )}{cos ( x ) + sin ( x )}

乘开

cos (x) - sin (x) + tan (x) (cos (x) + sin (x)) - (cos (x) + sin (x)) : - 2 sin (x) + tan (x) cos (x) + tan (x) sin (x)

cos (x) - sin (x) + tan (x) (cos (x) + sin (x)) - (cos (x) + sin (x))

乘开

tan (x) (cos (x) + sin (x)) : tan (x) cos (x) + tan (x) sin (x)

tan (x) (cos (x) + sin (x))

使用分配律:

a (b + c) = ab + a c

a = tan (x), b = cos (x), c = sin (x)

= tan (x) cos (x) + tan (x) sin (x)

= cos (x) - sin (x) + tan (x) cos (x) + tan (x) sin (x) - (cos (x) + sin (x))

- (cos (x) + sin (x)) : - cos (x) - sin (x)

- (cos (x) + sin (x))

打开括号

= - (cos (x)) - (sin (x))

使用加减运算法则

+ (- a) = - a

= - cos (x) - sin (x)

= cos (x) - sin (x) + tan (x) cos (x) + tan (x) sin (x) - cos (x) - sin (x)

化简

cos (x) - sin (x) + tan (x) cos (x) + tan (x) sin (x) - cos (x) - sin (x) : - 2 sin (x) + tan (x) cos (x) + tan (x) sin (x)

cos (x) - sin (x) + tan (x) cos (x) + tan (x) sin (x) - cos (x) - sin (x)

同类项相加:

cos (x) - cos (x) = 0

= - sin (x) + tan (x) cos (x) + tan (x) sin (x) - sin (x)

同类项相加:

- sin (x) - sin (x) = - 2 sin (x)

= - 2 sin (x) + tan (x) cos (x) + tan (x) sin (x)

= - 2 sin (x) + tan (x) cos (x) + tan (x) sin (x)

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

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

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

- 2 sin (x) + tan (x) cos (x) + tan (x) sin (x) = 0

使用三角恒等式改写

- 2 sin (x) + cos (x) tan (x) + sin (x) tan (x)

cos (x) tan (x) = sin (x)

cos (x) tan (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= cos (x) \frac{sin ( x )}{cos ( x )}

化简

cos (x) \frac{sin ( x )}{cos ( x )} : sin (x)

cos (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

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

= \frac{sin ( x ) cos ( x )}{cos ( x )}

约分:

cos (x)

= sin (x)

= sin (x)

= - 2 sin (x) + sin (x) + sin (x) tan (x)

化简

= - sin (x) + sin (x) tan (x)

- sin (x) + sin (x) tan (x) = 0

分解

- sin (x) + sin (x) tan (x) : sin (x) (tan (x) - 1)

- sin (x) + sin (x) tan (x)

因式分解出通项

sin (x)

= sin (x) (- 1 + tan (x))

sin (x) (tan (x) - 1) = 0

分别求解每个部分

sin (x) = 0 or tan (x) - 1 = 0

sin (x) = 0 : x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

sin (x) = 0

sin (x) = 0

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

x = 0 + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解

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

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

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

x = 36 0^{\circ} n

x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

tan (x) - 1 = 0 : x = 4 5^{\circ} + 18 0^{\circ} n

tan (x) - 1 = 0

将

1

到右边

tan (x) - 1 = 0

两边加上

1

tan (x) - 1 + 1 = 0 + 1

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 4 5^{\circ} + 18 0^{\circ} n

x = 4 5^{\circ} + 18 0^{\circ} n

合并所有解

x = 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

作图

流行的例子

tan(3B+5)=cot(2B+10)

tan (3 B + 5^{\circ}) = cot (2 B + 1 0^{\circ})

2cot(x)cos(x)+7=7csc(x)

2 cot (x) cos (x) + 7 = 7 csc (x)

2cos(2x-pi/3)=1

2 cos (2 x - \frac{π}{3}) = 1

cos(2θ)= 1/(sqrt(2))

cos (2 θ) = \frac{1}{2}

1 + 4 sin^{2} (θ) = 2