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受欢迎的三角函数 >

证明 tan(135+θ)=(-1+tan(θ))/(1+tan(θ))

解答

证明

tan (13 5^{\circ} + θ) = \frac{- 1 + tan ( θ )}{1 + tan ( θ )}

解答

真

求解步骤

tan (13 5^{\circ} + θ) = \frac{- 1 + tan ( θ )}{1 + tan ( θ )}

调整左侧

tan (13 5^{\circ} + θ)

使用三角恒等式改写

tan (13 5^{\circ} + θ)

使用基本三角恒等式:

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

= \frac{sin ( 13 5 ^{\circ} + θ )}{cos ( 13 5 ^{\circ} + θ )}

使用角和恒等式:

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

= \frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} + θ )}

使用角和恒等式:

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

= \frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

化简

\frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )} : - \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{sin ( 13 5 ^{\circ} ) cos ( θ ) + cos ( 13 5 ^{\circ} ) sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

sin (13 5^{\circ}) cos (θ) + cos (13 5^{\circ}) sin (θ) = \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

sin (13 5^{\circ}) cos (θ) + cos (13 5^{\circ}) sin (θ)

化简

sin (13 5^{\circ}) : \frac{2}{2}

sin (13 5^{\circ})

使用以下普通恒等式:

sin (13 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 (θ) + cos (13 5^{\circ}) sin (θ)

化简

cos (13 5^{\circ}) : - \frac{2}{2}

cos (13 5^{\circ})

使用以下普通恒等式:

cos (13 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 (θ) - \frac{2}{2} sin (θ)

= \frac{\frac{2}{2} cos ( θ ) - \frac{2}{2} sin ( θ )}{cos ( 13 5 ^{\circ} ) cos ( θ ) - sin ( 13 5 ^{\circ} ) sin ( θ )}

cos (13 5^{\circ}) cos (θ) - sin (13 5^{\circ}) sin (θ) = - \frac{2}{2} cos (θ) - \frac{2}{2} sin (θ)

cos (13 5^{\circ}) cos (θ) - sin (13 5^{\circ}) sin (θ)

化简

cos (13 5^{\circ}) : - \frac{2}{2}

cos (13 5^{\circ})

使用以下普通恒等式:

cos (13 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 (θ) - sin (13 5^{\circ}) sin (θ)

化简

sin (13 5^{\circ}) : \frac{2}{2}

sin (13 5^{\circ})

使用以下普通恒等式:

sin (13 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 (θ) - \frac{2}{2} sin (θ)

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

乘

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

\frac{2}{2} cos (θ)

分式相乘:

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

= \frac{2 cos ( θ )}{2}

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

乘

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

\frac{2}{2} sin (θ)

分式相乘:

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

= \frac{2 sin ( θ )}{2}

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

乘

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

\frac{2}{2} cos (θ)

分式相乘:

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

= \frac{2 cos ( θ )}{2}

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

乘

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

\frac{2}{2} sin (θ)

分式相乘:

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

= \frac{2 sin ( θ )}{2}

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

合并分式

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

使用法则

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

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

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

合并分式

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

使用法则

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

= \frac{2 cos ( θ ) - 2 sin ( θ )}{2}

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

分式相除:

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

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

约分:

2

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

因式分解出通项

2

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

因式分解出通项

2

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

约分:

2

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

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

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

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

化简

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

调整右侧

\frac{- 1 + tan ( θ )}{1 + tan ( θ )}

用 sin, cos 表示

\frac{- 1 + tan ( θ )}{1 + tan ( θ )}

使用基本三角恒等式:

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

= \frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

化简

\frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}} : \frac{- cos ( θ ) + sin ( θ )}{cos ( θ ) + sin ( θ )}

\frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{1 + \frac{s i n ( θ )}{c o s ( θ )}}

化简

1 + \frac{sin ( θ )}{cos ( θ )} : \frac{cos ( θ ) + sin ( θ )}{cos ( θ )}

1 + \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

= \frac{1 \cdot cos ( θ )}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

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{1 \cdot cos ( θ ) + sin ( θ )}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{- 1 + \frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

化简

- 1 + \frac{sin ( θ )}{cos ( θ )} : \frac{- cos ( θ ) + sin ( θ )}{cos ( θ )}

- 1 + \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

1 = \frac{1 cos ( θ )}{cos ( θ )}

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

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{- 1 \cdot cos ( θ ) + sin ( θ )}{cos ( θ )}

乘以:

1 \cdot cos (θ) = cos (θ)

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

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

分式相除:

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

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

约分:

cos (θ)

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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