zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

证明 tan(x)=((1-cos(2x)))/((sin(2x)))

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

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

化简

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

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

乘开

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

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

数字相除:

\frac{2}{2} = 1

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

约分:

sin (x)

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

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

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

使用基本三角恒等式:

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

= tan (x)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

证明 cot(pi-θ)=-cot(θ)

p ro v e cot (π - θ) = - cot (θ)

证明 sin(x+2pi)=sin(x)

p ro v e sin (x + 2 π) = sin (x)

证明 (cos(3a)cos(2a))/(sin(3a)sin(2a))=(cot(2a))/(tan(3a))

p ro v e \frac{cos ( 3 a ) cos ( 2 a )}{sin ( 3 a ) sin ( 2 a )} = \frac{cot ( 2 a )}{tan ( 3 a )}

证明 1/(cos^2(θ))=tan^2(θ)+1

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

证明 1/2 (cot(θ)-tan(θ))=cot(2θ)

p ro v e \frac{1}{2} (cot (θ) - tan (θ)) = cot (2 θ)