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受欢迎的三角函数 >

证明 (1+csc(x))(1-sin(x))=cot(x)cos(x)

解答

证明

(1 + csc (x)) (1 - sin (x)) = cot (x) cos (x)

解答

真

求解步骤

(1 + csc (x)) (1 - sin (x)) = cot (x) cos (x)

调整左侧

(1 + csc (x)) (1 - sin (x))

用 sin, cos 表示

(1 + csc (x)) (1 - sin (x))

使用基本三角恒等式:

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

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

化简

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

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

化简

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

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

将项转换为分式:

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

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

乘以:

1 \cdot sin (x) = sin (x)

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

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

乘开

(1 + sin (x)) (1 - sin (x)) : 1 - sin^{2} (x)

(1 + sin (x)) (1 - sin (x))

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (x)

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

= \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 )}

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

cos (x) cot (x)

cos (x) cot (x)

= cot (x) cos (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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