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受欢迎的三角函数 >

证明 cot(2x)+csc(2x)=cot(x)

解答

证明

cot (2 x) + csc (2 x) = cot (x)

解答

真

求解步骤

cot (2 x) + csc (2 x) = cot (x)

调整左侧

cot (2 x) + csc (2 x)

用 sin, cos 表示

cot (2 x) + csc (2 x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

使用法则

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

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

使用倍角公式:

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

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

化简

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

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

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

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

对同类项分组

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

1 - 1 = 0

= 2 cos^{2} (x)

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

数字相除:

\frac{2}{2} = 1

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

约分:

cos (x)

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

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

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

使用基本三角恒等式:

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

= cot (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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