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受欢迎的三角函数 >

证明 csc(-x)-sin(-x)=cos(-x)cot(-x)

解答

证明

csc (- x) - sin (- x) = cos (- x) cot (- x)

解答

真

求解步骤

csc (- x) - sin (- x) = cos (- x) cot (- x)

调整左侧

csc (- x) - sin (- x)

使用负角恒等式:

sin (- x) = - sin (x)

= csc (- x) - (- sin (x))

使用负角恒等式:

csc (- x) = - csc (x)

= - (- sin (x)) - csc (x)

用 sin, cos 表示

- (- sin (x)) - csc (x)

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

= - (- sin (x)) - \frac{1}{sin ( x )}

化简

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

- (- sin (x)) - \frac{1}{sin ( x )}

使用法则

- (- a) = a

= sin (x) - \frac{1}{sin ( x )}

将项转换为分式:

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

= \frac{sin ( x ) sin ( x )}{sin ( x )} - \frac{1}{sin ( 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{sin ( x ) sin ( x ) - 1}{sin ( x )}

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

sin (x) sin (x) - 1

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

= sin^{2} (x) - 1

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

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

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

调整右侧

cos (- x) cot (- x)

使用负角恒等式:

cos (- x) = cos (x)

= cos (x) cot (- x)

使用负角恒等式:

cot (- x) = - cot (x)

= cos (x) (- cot (x))

用 sin, cos 表示

(- cot (x)) cos (x)

使用基本三角恒等式:

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

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

化简

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

(- \frac{cos ( x )}{sin ( x )}) cos (x)

去除括号:

(- a) = - a

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

分式相乘:

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

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

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

cos (x) cos (x)

使用指数法则:

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

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

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

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

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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