证明 csc(2x)+cot(2x)=(1+cos(2x))/(sin(2x))

解答

证明

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

真

求解步骤

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

调整左侧

csc (2 x) + cot (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 )}

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

p ro v e 1 - 2 sin^{2} (t) = 2 cos^{2} (t) - 1

p ro v e \frac{1}{cos ^{2} ( θ )} = sec^{2} (θ)

p ro v e - cos (2 t) sin (2 t) + sin (2 t) cos (2 t) + 0 = 0

p ro v e \frac{tan ( θ ) sin ( θ )}{sec ( θ ) - 1} = 1 + cos (θ)

p ro v e 1 - 2 sin^{2} (x) = - 1 + cos^{2} (x)