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受欢迎的三角函数 >

证明 (sin^2(x))/(cos^2(x))+1=sec^2(x)

解答

证明

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

解答

真

求解步骤

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

调整右侧

sec^{2} (x)

使用三角恒等式改写

sec^{2} (x)

使用毕达哥拉斯恒等式:

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

= tan^{2} (x) + 1

= tan^{2} (x) + 1

用 sin, cos 表示

1 + tan^{2} (x)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

= 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 )}

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

乘开

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

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

使用分式法则:

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

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

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

使用法则

\frac{a}{a} = 1

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

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

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

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

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

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p ro v e csc (\frac{π}{12}) = sec (\frac{5 π}{12})

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p ro v e 1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )} = sin^{2} (θ)

证明 csc^2(t)tan^2(t)-1=tan^2(t)

p ro v e csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)