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受欢迎的三角函数 >

证明 tan(-x)tan(pi/2-x)=-1

解答

证明

tan (- x) tan (\frac{π}{2} - x) = - 1

解答

真

求解步骤

tan (- x) tan (\frac{π}{2} - x) = - 1

调整左侧

tan (- x) tan (\frac{π}{2} - x)

使用负角恒等式:

tan (- x) = - tan (x)

= (- tan (x)) tan (\frac{π}{2} - x)

使用三角恒等式改写

tan (\frac{π}{2} - x)

使用基本三角恒等式:

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

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

使用角差恒等式:

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

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

使用角差恒等式:

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

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

化简

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

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

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

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

sin (\frac{π}{2}) cos (x) = cos (x)

sin (\frac{π}{2}) cos (x)

化简

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot cos (x)

乘以:

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

= cos (x)

cos (\frac{π}{2}) sin (x) = 0

cos (\frac{π}{2}) sin (x)

化简

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

= cos (x) - 0

cos (x) - 0 = cos (x)

= cos (x)

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

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

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

cos (\frac{π}{2}) cos (x) = 0

cos (\frac{π}{2}) cos (x)

化简

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

使用以下普通恒等式:

cos (\frac{π}{2}) = 0

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}

= 0

= 0 \cdot cos (x)

使用法则

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (x) = sin (x)

sin (\frac{π}{2}) sin (x)

化简

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

使用以下普通恒等式:

sin (\frac{π}{2}) = 1

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}

= 1

= 1 \cdot sin (x)

乘以:

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

= sin (x)

= 0 + sin (x)

0 + sin (x) = sin (x)

= sin (x)

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

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

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

化简

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

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

去除括号:

(- a) = - a

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

分式相乘:

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

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

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

- \frac{cos ( x ) \frac{s i n ( x )}{c o s ( x )}}{sin ( x )} = - 1

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

乘

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)

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

使用法则

\frac{a}{a} = 1

= - 1

= - 1

= - 1

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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