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受欢迎的三角函数 >

证明 tan(x)=((1-cos(2x)))/(sin(2x))

解答

证明

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

解答

真

求解步骤

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

调整右侧

\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

流行的例子

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p ro v e sin^{2} (a) - cos^{2} (a) = 1 - 2 cos^{2} (a)

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p ro v e tan (2 A) - 2 tan (2 A) sin^{2} (A) = sin (2 A)

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p ro v e sin^{2} (x) = \frac{1}{csc ^{2} ( x )}

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p ro v e \frac{cot ( 4 u ) - 1}{cot ( 4 u ) + 1} = \frac{1 - tan ( 4 u )}{1 + tan ( 4 u )}

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p ro v e sin (- x) = sin (x - π)