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受欢迎的三角函数 >

证明 (sin(x))/(1-cos(-x))=csc(x)+cot(x)

解答

证明

\frac{sin ( x )}{1 - cos ( - x )} = csc (x) + cot (x)

解答

真

求解步骤

\frac{sin ( x )}{1 - cos ( - x )} = csc (x) + cot (x)

调整左侧

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

使用负角恒等式:

cos (- x) = cos (x)

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

使用三角恒等式改写

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

乘以

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

= \frac{sin ( x )}{\frac{( 1 - c o s ( x ) ) ( 1 + c o s ( x ) )}{1 + c o s ( x )}}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

使用毕达哥拉斯恒等式:

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

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

= \frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

化简

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}} : \frac{1 + cos ( x )}{sin ( x )}

\frac{sin ( x )}{\frac{s i n ^{2} ( x )}{1 + c o s ( x )}}

使用分式法则:

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

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

约分:

sin (x)

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

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

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

调整右侧

csc (x) + cot (x)

用 sin, cos 表示

cot (x) + csc (x)

使用基本三角恒等式:

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

= \frac{cos ( x )}{sin ( x )} + csc (x)

使用基本三角恒等式:

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

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

化简

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

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

使用法则

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

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

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

= \frac{1 + cos ( 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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