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受欢迎的三角函数 >

证明 1-tan^2(x/2)=(2cos(x))/(1+cos(x))

解答

证明

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

解答

真

求解步骤

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

令

: u = \frac{x}{2}

1 - tan^{2} (u) = \frac{2 cos ( 2 u )}{1 + cos ( 2 u )}

证明

1 - tan^{2} (u) = \frac{2 cos ( 2 u )}{1 + cos ( 2 u )} :

真

1 - tan^{2} (u) = \frac{2 cos ( 2 u )}{1 + cos ( 2 u )}

调整左侧

1 - tan^{2} (u)

用 sin, cos 表示

1 - tan^{2} (u)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

将项转换为分式:

1 = \frac{1 cos ^{2} ( u )}{cos ^{2} ( u )}

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

因为分母相等，所以合并分式:

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

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

乘以:

1 \cdot cos^{2} (u) = cos^{2} (u)

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

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

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

使用三角恒等式改写

\frac{cos ^{2} ( u ) - sin ^{2} ( u )}{cos ^{2} ( u )}

使用毕达哥拉斯恒等式:

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

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

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

乘开

cos^{2} (u) - (1 - cos^{2} (u)) : 2 cos^{2} (u) - 1

cos^{2} (u) - (1 - cos^{2} (u))

- (1 - cos^{2} (u)) : - 1 + cos^{2} (u)

- (1 - cos^{2} (u))

打开括号

= - (1) - (- cos^{2} (u))

使用加减运算法则

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

= - 1 + cos^{2} (u)

= cos^{2} (u) - 1 + cos^{2} (u)

化简

cos^{2} (u) - 1 + cos^{2} (u) : 2 cos^{2} (u) - 1

cos^{2} (u) - 1 + cos^{2} (u)

对同类项分组

= cos^{2} (u) + cos^{2} (u) - 1

同类项相加:

cos^{2} (u) + cos^{2} (u) = 2 cos^{2} (u)

= 2 cos^{2} (u) - 1

= 2 cos^{2} (u) - 1

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

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

调整右侧

\frac{2 cos ( 2 u )}{1 + cos ( 2 u )}

使用三角恒等式改写

\frac{2 cos ( 2 u )}{1 + cos ( 2 u )}

使用倍角公式:

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

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

化简

\frac{2 ( 2 cos ^{2} ( u ) - 1 )}{1 + 2 cos ^{2} ( u ) - 1} : \frac{2 cos ^{2} ( u ) - 1}{cos ^{2} ( u )}

\frac{2 ( 2 cos ^{2} ( u ) - 1 )}{1 + 2 cos ^{2} ( u ) - 1}

1 + 2 cos^{2} (u) - 1 = 2 cos^{2} (u)

1 + 2 cos^{2} (u) - 1

对同类项分组

= 2 cos^{2} (u) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (u)

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

数字相除:

\frac{2}{2} = 1

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

因此

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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