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受欢迎的三角函数 >

证明 ((cos^2(x)+4cos(x)+4))/((cos(x)+2))=((2sec(x)+1))/(sec(x))

解答

证明

\frac{( cos ^{2} ( x ) + 4 cos ( x ) + 4 )}{( cos ( x ) + 2 )} = \frac{( 2 sec ( x ) + 1 )}{sec ( x )}

解答

真

求解步骤

\frac{cos ^{2} ( x ) + 4 cos ( x ) + 4}{cos ( x ) + 2} = \frac{2 sec ( x ) + 1}{sec ( x )}

调整左侧

\frac{cos ^{2} ( x ) + 4 cos ( x ) + 4}{cos ( x ) + 2}

化简

\frac{4 + cos ^{2} ( x ) + 4 cos ( x )}{2 + cos ( x )} : cos (x) + 2

\frac{4 + cos ^{2} ( x ) + 4 cos ( x )}{2 + cos ( x )}

分解

4 + cos^{2} (x) + 4 cos (x) : (cos (x) + 2)^{2}

4 + cos^{2} (x) + 4 cos (x)

改写成标准形式

a x^{2} + b x + c

= cos^{2} (x) + 4 cos (x) + 4

将

cos^{2} (x) + 4 cos (x) + 4

改写为

cos^{2} (x) + 2 cos (x) \cdot 2 + 2^{2}

cos^{2} (x) + 4 cos (x) + 4

将

4

改写为

2^{2}

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

将

2^{2} cos (x)

改写为

2 cos (x) \cdot 2

= cos^{2} (x) + 2 cos (x) \cdot 2 + 2^{2}

= cos^{2} (x) + 2 cos (x) \cdot 2 + 2^{2}

使用完全平方公式:

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

a = cos (x), b = 2

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

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

约分:

cos (x) + 2

= cos (x) + 2

= 2 + cos (x)

调整右侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

使用分式法则:

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

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

使用分式法则:

\frac{a}{1} = a

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

分式相乘:

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

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

约分:

cos (x)

= cos (x) + 2

= cos (x) + 2

= 2 + cos (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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