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受欢迎的三角函数 >

证明 cos(x)(1+tan(x))^2=sec(x)+2sin(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

cos (x) (1 + tan (x))^{2}

用 sin, cos 表示

(1 + tan (x))^{2} cos (x)

使用基本三角恒等式:

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

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

化简

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

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

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

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

化简

1 + \frac{sin ( x )}{cos ( x )} : \frac{cos ( x ) + sin ( x )}{cos ( x )}

1 + \frac{sin ( x )}{cos ( x )}

将项转换为分式:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{sin ( x )}{cos ( 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{1 \cdot cos ( x ) + sin ( x )}{cos ( x )}

乘以:

1 \cdot cos (x) = cos (x)

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

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

使用指数法则:

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

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

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

分式相乘:

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

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

约分:

cos (x)

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

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

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

乘开

(cos (x) + sin (x))^{2} : cos^{2} (x) + 2 cos (x) sin (x) + sin^{2} (x)

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

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = cos (x), b = sin (x)

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= \frac{1 + 2 cos ( x ) sin ( x )}{cos ( 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 + 2 \cdot \frac{1}{s e c ( x )} sin ( x )}{\frac{1}{s e c ( x )}}

化简

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

使用分式法则:

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

= \frac{( 1 + 2 \cdot \frac{1}{s e c ( x )} sin ( x ) ) sec ( x )}{1}

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

2 \cdot \frac{1}{sec ( x )} sin (x)

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

= \frac{sec ( x ) ( \frac{2 s i n ( x )}{s e c ( x )} + 1 )}{1}

化简

1 + \frac{2 sin ( x )}{sec ( x )} : \frac{sec ( x ) + 2 sin ( x )}{sec ( x )}

1 + \frac{2 sin ( x )}{sec ( x )}

将项转换为分式:

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

= \frac{1 \cdot sec ( x )}{sec ( x )} + \frac{2 sin ( x )}{sec ( 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{1 \cdot sec ( x ) + 2 sin ( x )}{sec ( x )}

乘以:

1 \cdot sec (x) = sec (x)

= \frac{sec ( x ) + 2 sin ( x )}{sec ( x )}

= \frac{\frac{s e c ( x ) + 2 s i n ( x )}{s e c ( x )} sec ( x )}{1}

使用分式法则:

\frac{a}{1} = a

= \frac{sec ( x ) + 2 sin ( x )}{sec ( x )} sec (x)

分式相乘:

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

= \frac{( sec ( x ) + 2 sin ( x ) ) sec ( x )}{sec ( x )}

约分:

sec (x)

= sec (x) + 2 sin (x)

sec (x) + 2 sin (x)

sec (x) + 2 sin (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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