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 1/(1+sin(x))+1/(1-sin(x))=2+2tan^2(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

化简

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

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

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

的最小公倍数:

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

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

最小公倍数 (LCM)

计算出由出现在

1 + sin (x)

或

1 - sin (x)

中的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

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

对于

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

将分母和分子乘以

- sin (x) + 1

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

对于

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

将分母和分子乘以

sin (x) + 1

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

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

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

- sin (x) + 1 + sin (x) + 1 = 2

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

对同类项分组

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

同类项相加:

- sin (x) + sin (x) = 0

= 1 + 1

数字相加:

1 + 1 = 2

= 2

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

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

使用三角恒等式改写

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

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

调整右侧

2 + 2 tan^{2} (x)

分解

2 + 2 tan^{2} (x) : 2 (1 + tan^{2} (x))

2 + 2 tan^{2} (x)

改写为

= 2 \cdot 1 + 2 tan^{2} (x)

因式分解出通项

2

= 2 (1 + tan^{2} (x))

= (1 + tan^{2} (x)) \cdot 2

用 sin, cos 表示

(1 + tan^{2} (x)) \cdot 2

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

化简

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

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

将项转换为分式:

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

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

乘以:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

化简

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

= \frac{2}{cos ^{2} ( 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 sec(2x)=(sec^2(x))/(1-tan^2(x))

p ro v e sec (2 x) = \frac{sec ^{2} ( x )}{1 - tan ^{2} ( x )}

证明 (sin(3x)-sin(5x))/(2cos(4x))=-sin(x)

p ro v e \frac{sin ( 3 x ) - sin ( 5 x )}{2 cos ( 4 x )} = - sin (x)

证明 tan(a)+cot(a)=2csc(2a)

p ro v e tan (a) + cot (a) = 2 csc (2 a)

证明 (sec(t)-1)/(tan(t))=(tan(t))/(sec(t)+1)

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

证明 3cos^2(x)-3sin^2(x)=3-6sin^2(x)

p ro v e 3 cos^{2} (x) - 3 sin^{2} (x) = 3 - 6 sin^{2} (x)