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受欢迎的三角函数 >

证明 sec(2x)=((sec^2(x)))/(2-sec^2(x))

解答

证明

sec (2 x) = \frac{( sec ^{2} ( x ) )}{2 - sec ^{2} ( x )}

解答

真

求解步骤

sec (2 x) = \frac{sec ^{2} ( x )}{2 - sec ^{2} ( x )}

调整右侧

\frac{sec ^{2} ( x )}{2 - sec ^{2} ( x )}

用 sin, cos 表示

\frac{sec ^{2} ( x )}{2 - sec ^{2} ( x )}

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= \frac{( \frac{1}{c o s ( x )} ) ^{2}}{2 - ( \frac{1}{c o s ( x )} ) ^{2}}

化简

\frac{( \frac{1}{c o s ( x )} ) ^{2}}{2 - ( \frac{1}{c o s ( x )} ) ^{2}} : \frac{1}{2 cos ^{2} ( x ) - 1}

\frac{( \frac{1}{c o s ( x )} ) ^{2}}{2 - ( \frac{1}{c o s ( x )} ) ^{2}}

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

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{( \frac{1}{c o s ( x )} ) ^{2}}{2 - \frac{1}{c o s ^{2} ( x )}}

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

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{\frac{1}{c o s ^{2} ( x )}}{2 - \frac{1}{c o s ^{2} ( x )}}

使用分式法则:

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

= \frac{1}{cos ^{2} ( x ) ( 2 - \frac{1}{c o s ^{2} ( x )} )}

化简

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

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

将项转换为分式:

2 = \frac{2 cos ^{2} ( x )}{cos ^{2} ( x )}

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

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

= \frac{1}{\frac{2 c o s ^{2} ( x ) - 1}{c o s ^{2} ( x )} cos ^{2} ( x )}

乘

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

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

分式相乘:

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

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

约分:

cos^{2} (x)

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

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

使用三角恒等式改写

使用基本三角恒等式:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{\frac{1}{s e c ( 2 x )}}

化简

\frac{1}{\frac{1}{s e c ( 2 x )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ( 2 x )}{1}

使用法则

\frac{a}{1} = a

= sec (2 x)

sec (2 x)

sec (2 x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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