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受欢迎的三角函数 >

证明 tan(pi/4-θ)=(cos(θ)-sin(θ))/(cos(θ)+sin(θ))

解答

证明

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

解答

真

求解步骤

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

调整左侧

tan (\frac{π}{4} - θ)

使用三角恒等式改写

tan (\frac{π}{4} - θ)

使用基本三角恒等式:

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

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

使用角差恒等式:

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

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

使用角差恒等式:

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

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

化简

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

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

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

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

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 (θ) - cos (\frac{π}{4}) sin (θ)

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 (θ) - \frac{2}{2} sin (θ)

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

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

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

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 (θ) + sin (\frac{π}{4}) sin (θ)

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 (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 )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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