证明 (1-sin(2x))/(cos(2x))=(cos(2x))/(1+sin(2x))

解答

证明

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

真

求解步骤

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

调整左侧

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

使用三角恒等式改写

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

乘以

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

cos (2 x)

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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