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受欢迎的三角函数 >

证明 cos(2α)=-1+2/(1+tan^2(α))

解答

证明

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

解答

真

求解步骤

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

调整右侧

- 1 + \frac{2}{1 + tan ^{2} ( α )}

用 sin, cos 表示

- 1 + \frac{2}{1 + tan ^{2} ( α )}

使用基本三角恒等式:

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

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

化简

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

- 1 + \frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}}

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

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} ( α ) + sin ^{2} ( α )}{cos ^{2} ( α )}

乘以:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

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

使用分式法则:

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

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

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

将项转换为分式:

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

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

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} ( α ) + sin ^{2} ( α ) ) + 2 cos ^{2} ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

乘以:

1 \cdot (cos^{2} (α) + sin^{2} (α)) = (cos^{2} (α) + sin^{2} (α))

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

乘开

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α) : cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) : - cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α))

打开括号

= - (cos^{2} (α)) - (sin^{2} (α))

使用加减运算法则

+ (- a) = - a

= - cos^{2} (α) - sin^{2} (α)

= - cos^{2} (α) - sin^{2} (α) + 2 cos^{2} (α)

同类项相加:

- cos^{2} (α) + 2 cos^{2} (α) = cos^{2} (α)

= cos^{2} (α) - sin^{2} (α)

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

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

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

使用三角恒等式改写

\frac{cos ^{2} ( α ) - sin ^{2} ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

使用毕达哥拉斯恒等式:

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

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

使用法则

\frac{a}{1} = a

= cos^{2} (α) - sin^{2} (α)

使用倍角公式:

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

= cos (2 α)

= cos (2 α)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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