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受欢迎的三角函数 >

证明 (1+sec(θ))/(sec(θ))=2cos^2(θ/2)

解答

证明

\frac{1 + sec ( θ )}{sec ( θ )} = 2 cos^{2} (\frac{θ}{2})

解答

真

求解步骤

\frac{1 + sec ( θ )}{sec ( θ )} = 2 cos^{2} (\frac{θ}{2})

令

: u = \frac{θ}{2}

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

证明

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

真

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

\frac{1 + \frac{1}{c o s ( 2 u )}}{\frac{1}{c o s ( 2 u )}} : cos (2 u) + 1

\frac{1 + \frac{1}{c o s ( 2 u )}}{\frac{1}{c o s ( 2 u )}}

使用分式法则:

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

= \frac{( 1 + \frac{1}{c o s ( 2 u )} ) cos ( 2 u )}{1}

化简

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

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

乘以:

1 \cdot cos (2 u) = cos (2 u)

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

= \frac{\frac{c o s ( 2 u ) + 1}{c o s ( 2 u )} cos ( 2 u )}{1}

使用分式法则:

\frac{a}{1} = a

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

分式相乘:

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

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

约分:

cos (2 u)

= cos (2 u) + 1

= 1 + cos (2 u)

= 1 + cos (2 u)

使用三角恒等式改写

1 + cos (2 u)

使用倍角公式:

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

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

化简

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)

= 2 cos^{2} (u)

= 2 cos^{2} (u)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

因此

\frac{1 + sec ( θ )}{sec ( θ )} = 2 cos^{2} (\frac{θ}{2})

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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