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受欢迎的三角函数 >

证明 cos(a)(sec(a)-cos(a))=sin^2(a)

解答

证明

cos (a) (sec (a) - cos (a)) = sin^{2} (a)

解答

真

求解步骤

cos (a) (sec (a) - cos (a)) = sin^{2} (a)

调整左侧

cos (a) (sec (a) - cos (a))

用 sin, cos 表示

(- cos (a) + sec (a)) cos (a)

使用基本三角恒等式:

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

= (- cos (a) + \frac{1}{cos ( a )}) cos (a)

化简

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

(- cos (a) + \frac{1}{cos ( a )}) cos (a)

化简

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

- cos (a) + \frac{1}{cos ( a )}

将项转换为分式:

cos (a) = \frac{cos ( a ) cos ( a )}{cos ( a )}

= - \frac{cos ( a ) cos ( a )}{cos ( a )} + \frac{1}{cos ( a )}

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{- cos ( a ) cos ( a ) + 1}{cos ( a )}

- cos (a) cos (a) + 1 = - cos^{2} (a) + 1

- cos (a) cos (a) + 1

cos (a) cos (a) = cos^{2} (a)

cos (a) cos (a)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (a) cos (a) = cos^{1 + 1} (a)

= cos^{1 + 1} (a)

数字相加:

1 + 1 = 2

= cos^{2} (a)

= - cos^{2} (a) + 1

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

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

分式相乘:

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

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

约分:

cos (a)

= - - cos^{2} (a) + 1

= - cos^{2} (a) + 1

= 1 - cos^{2} (a)

使用三角恒等式改写

1 - cos^{2} (a)

使用毕达哥拉斯恒等式:

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

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

= sin^{2} (a)

= sin^{2} (a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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