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受欢迎的三角函数 >

证明 (cos(2x))/(1-sin^2(x))=2-sec^2(x)

解答

证明

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

解答

真

求解步骤

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

调整右侧

2 - sec^{2} (x)

用 sin, cos 表示

2 - sec^{2} (x)

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

将项转换为分式:

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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