zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 tan(pi/4-t)=(1-tan(t))/(1+tan(t))

解答

证明

tan (\frac{π}{4} - t) = \frac{1 - tan ( t )}{1 + tan ( t )}

解答

真

求解步骤

tan (\frac{π}{4} - t) = \frac{1 - tan ( t )}{1 + tan ( t )}

调整左侧

tan (\frac{π}{4} - t)

使用三角恒等式改写

tan (\frac{π}{4} - t)

使用基本三角恒等式:

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

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

使用角差恒等式:

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

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

使用角差恒等式:

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

= \frac{sin ( \frac{π}{4} ) cos ( t ) - cos ( \frac{π}{4} ) sin ( t )}{cos ( \frac{π}{4} ) cos ( t ) + sin ( \frac{π}{4} ) sin ( t )}

化简

\frac{sin ( \frac{π}{4} ) cos ( t ) - cos ( \frac{π}{4} ) sin ( t )}{cos ( \frac{π}{4} ) cos ( t ) + sin ( \frac{π}{4} ) sin ( t )} : \frac{cos ( t ) - sin ( t )}{cos ( t ) + sin ( t )}

\frac{sin ( \frac{π}{4} ) cos ( t ) - cos ( \frac{π}{4} ) sin ( t )}{cos ( \frac{π}{4} ) cos ( t ) + sin ( \frac{π}{4} ) sin ( t )}

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

sin (\frac{π}{4}) cos (t) - cos (\frac{π}{4}) sin (t)

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

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

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} 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 (t) - \frac{2}{2} sin (t)

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

cos (\frac{π}{4}) cos (t) + sin (\frac{π}{4}) sin (t) = \frac{2}{2} cos (t) + \frac{2}{2} sin (t)

cos (\frac{π}{4}) cos (t) + sin (\frac{π}{4}) sin (t)

化简

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{2}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} 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 (t) + sin (\frac{π}{4}) sin (t)

化简

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{2}{2}

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}

= \frac{2}{2}

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

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

乘

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

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

分式相乘:

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

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

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

乘

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

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

分式相乘:

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

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

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

乘

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

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

分式相乘:

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

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

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

乘

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

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

分式相乘:

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

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

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

合并分式

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

使用法则

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

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

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

合并分式

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

使用法则

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

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

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

分式相除:

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

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

约分:

2

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

因式分解出通项

2

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

因式分解出通项

2

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

约分:

2

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

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

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

调整右侧

\frac{1 - tan ( t )}{1 + tan ( t )}

用 sin, cos 表示

\frac{1 - tan ( t )}{1 + tan ( t )}

使用基本三角恒等式:

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

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

化简

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

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

化简

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

1 + \frac{sin ( t )}{cos ( t )}

将项转换为分式:

1 = \frac{1 cos ( t )}{cos ( t )}

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

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 ( t ) + sin ( t )}{cos ( t )}

乘以:

1 \cdot cos (t) = cos (t)

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

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

化简

1 - \frac{sin ( t )}{cos ( t )} : \frac{cos ( t ) - sin ( t )}{cos ( t )}

1 - \frac{sin ( t )}{cos ( t )}

将项转换为分式:

1 = \frac{1 cos ( t )}{cos ( t )}

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

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 ( t ) - sin ( t )}{cos ( t )}

乘以:

1 \cdot cos (t) = cos (t)

= \frac{cos ( t ) - sin ( t )}{cos ( t )}

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

分式相除:

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

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

约分:

cos (t)

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明-tan(x)=tan(-x)

p ro v e - tan (x) = tan (- x)

证明 sin(3θ)=3cos^2(θ)sin(θ)-sin^3(θ)

p ro v e sin (3 θ) = 3 cos^{2} (θ) sin (θ) - sin^{3} (θ)

证明 sin(x)sin(3x)=4sin(x)cos^2(x)

p ro v e sin (x) sin (3 x) = 4 sin (x) cos^{2} (x)

证明 sec(pi/2-t)=csc(t)

p ro v e sec (\frac{π}{2} - t) = csc (t)

证明 sin(3u)=sin(u)(3-4sin^2(u))

p ro v e sin (3 u) = sin (u) (3 - 4 sin^{2} (u))