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受欢迎的三角函数 >

证明 tan(x)=csc(2x)-cot(2x)

解答

证明

tan (x) = csc (2 x) - cot (2 x)

解答

真

求解步骤

tan (x) = csc (2 x) - cot (2 x)

调整右侧

csc (2 x) - cot (2 x)

用 sin, cos 表示

- cot (2 x) + csc (2 x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= - \frac{cos ( 2 x )}{sin ( 2 x )} + \frac{1}{sin ( 2 x )}

化简

- \frac{cos ( 2 x )}{sin ( 2 x )} + \frac{1}{sin ( 2 x )} : \frac{- cos ( 2 x ) + 1}{sin ( 2 x )}

- \frac{cos ( 2 x )}{sin ( 2 x )} + \frac{1}{sin ( 2 x )}

使用法则

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

= \frac{- cos ( 2 x ) + 1}{sin ( 2 x )}

= \frac{- cos ( 2 x ) + 1}{sin ( 2 x )}

= \frac{1 - cos ( 2 x )}{sin ( 2 x )}

使用三角恒等式改写

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

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= \frac{1 - cos ( 2 x )}{2 sin ( x ) cos ( x )}

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= \frac{1 - ( 1 - 2 sin ^{2} ( x ) )}{2 sin ( x ) cos ( x )}

化简

\frac{1 - ( 1 - 2 sin ^{2} ( x ) )}{2 sin ( x ) cos ( x )} : \frac{sin ( x )}{cos ( x )}

\frac{1 - ( 1 - 2 sin ^{2} ( x ) )}{2 sin ( x ) cos ( x )}

乘开

1 - (1 - 2 sin^{2} (x)) : 2 sin^{2} (x)

1 - (1 - 2 sin^{2} (x))

- (1 - 2 sin^{2} (x)) : - 1 + 2 sin^{2} (x)

- (1 - 2 sin^{2} (x))

打开括号

= - (1) - (- 2 sin^{2} (x))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (x)

= 1 - 1 + 2 sin^{2} (x)

1 - 1 = 0

= 2 sin^{2} (x)

= \frac{2 sin ^{2} ( x )}{2 sin ( x ) cos ( x )}

数字相除:

\frac{2}{2} = 1

= \frac{sin ^{2} ( x )}{sin ( x ) cos ( x )}

约分:

sin (x)

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

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

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

使用基本三角恒等式:

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

= tan (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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