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受欢迎的三角函数 >

证明 sin(α)=(2tan(α/2))/(1+tan^2(α/2))

解答

证明

sin (α) = \frac{2 tan ( \frac{α}{2} )}{1 + tan ^{2} ( \frac{α}{2} )}

解答

真

求解步骤

sin (α) = \frac{2 tan ( \frac{α}{2} )}{1 + tan ^{2} ( \frac{α}{2} )}

令

: u = \frac{α}{2}

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

证明

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

真

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

调整右侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

= \frac{2 \cdot \frac{s i n ( u )}{c o s ( u )}}{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}

化简

\frac{2 \cdot \frac{s i n ( u )}{c o s ( u )}}{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}} : \frac{2 sin ( u ) cos ( u )}{cos ^{2} ( u ) + sin ^{2} ( u )}

\frac{2 \cdot \frac{s i n ( u )}{c o s ( u )}}{1 + ( \frac{s i n ( u )}{c o s ( u )} ) ^{2}}

使用指数法则:

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

= \frac{2 \cdot \frac{s i n ( u )}{c o s ( u )}}{1 + \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}

乘

2 \cdot \frac{sin ( u )}{cos ( u )} : \frac{2 sin ( u )}{cos ( u )}

2 \cdot \frac{sin ( u )}{cos ( u )}

分式相乘:

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

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

= \frac{\frac{2 s i n ( u )}{c o s ( u )}}{1 + \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )}}

使用分式法则:

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

= \frac{sin ( u ) \cdot 2}{cos ( u ) ( 1 + \frac{s i n ^{2} ( u )}{c o s ^{2} ( u )} )}

化简

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

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{2 sin ( u )}{\frac{c o s ^{2} ( u ) + s i n ^{2} ( u )}{c o s ^{2} ( u )} cos ( u )}

乘

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

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

分式相乘:

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

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

约分:

cos (u)

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

= \frac{2 sin ( u )}{\frac{c o s ^{2} ( u ) + s i n ^{2} ( u )}{c o s ( u )}}

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

= \frac{sin ( 2 u )}{1}

使用法则

\frac{a}{1} = a

= sin (2 u)

= sin (2 u)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

因此

sin (α) = \frac{2 tan ( \frac{α}{2} )}{1 + tan ^{2} ( \frac{α}{2} )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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