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受欢迎的三角函数 >

证明 cot(a)=tan(pi/2-a)

解答

证明

cot (a) = tan (\frac{π}{2} - a)

解答

真

求解步骤

cot (a) = tan (\frac{π}{2} - a)

调整右侧

tan (\frac{π}{2} - a)

使用三角恒等式改写

tan (\frac{π}{2} - a)

使用基本三角恒等式:

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

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

使用角差恒等式:

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

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

使用角差恒等式:

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

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

化简

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

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

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

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

sin (\frac{π}{2}) cos (a) = cos (a)

sin (\frac{π}{2}) cos (a)

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 (a)

乘以:

1 \cdot cos (a) = cos (a)

= cos (a)

cos (\frac{π}{2}) sin (a) = 0

cos (\frac{π}{2}) sin (a)

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 (a)

使用法则

0 \cdot a = 0

= 0

= cos (a) - 0

cos (a) - 0 = cos (a)

= cos (a)

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

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

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

cos (\frac{π}{2}) cos (a) = 0

cos (\frac{π}{2}) cos (a)

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 (a)

使用法则

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (a) = sin (a)

sin (\frac{π}{2}) sin (a)

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 (a)

乘以:

1 \cdot sin (a) = sin (a)

= sin (a)

= 0 + sin (a)

0 + sin (a) = sin (a)

= sin (a)

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

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

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

使用基本三角恒等式:

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

= cot (a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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