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受欢迎的三角函数 >

证明 (1+tan^2(x))cos(x)=sec(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

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

分式相乘:

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

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

约分:

cos (x)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

使用分式法则:

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

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

使用法则

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e \frac{2}{cos ^{2} ( x )} = 2 sec^{2} (x)

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p ro v e 2 cos^{2} (a) = (1 + sec (2 a)) cos (2 a)

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p ro v e sin (\frac{13 π}{2} + x) = cos (x)

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p ro v e tan^{2} (- θ) + 1 = sec^{2} (θ)

证明 (1+csc(β))/(cot(β)+cos(β))=sec(β)

p ro v e \frac{1 + csc ( β )}{cot ( β ) + cos ( β )} = sec (β)