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(θ))=sec^2(θ)+sec(θ)tan(θ)

解答

证明

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

解答

真

求解步骤

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

调整右侧

sec^{2} (θ) + sec (θ) tan (θ)

用 sin, cos 表示

sec^{2} (θ) + sec (θ) tan (θ)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= (\frac{1}{cos ( θ )})^{2} + \frac{1}{cos ( θ )} tan (θ)

使用基本三角恒等式:

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

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

化简

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

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

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

(\frac{1}{cos ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

\frac{1}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{1 \cdot sin ( θ )}{cos ( θ ) cos ( θ )}

乘以:

1 \cdot sin (θ) = sin (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= \frac{sin ( θ )}{cos ^{2} ( θ )}

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

使用法则

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

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

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

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

使用三角恒等式改写

\frac{1 + sin ( θ )}{cos ^{2} ( θ )}

使用毕达哥拉斯恒等式:

cos^{2} (θ) + sin^{2} (θ) = 1

cos^{2} (θ) = 1 - sin^{2} (θ)

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

分解

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

1 - sin^{2} (θ)

使用平方差公式:

θ^{2} - y^{2} = (θ + y) (θ - y)

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

= (1 + sin (θ)) (1 - sin (θ))

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

约分:

1 + sin (θ)

= \frac{1}{1 - sin ( θ )}

= \frac{1}{1 - sin ( θ )}

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 2sin(x)=tan(x)

p ro v e 2 sin (x) = tan (x)

证明 cot(x)-1=csc(x)(cos(x)-sin(x))

p ro v e cot (x) - 1 = csc (x) (cos (x) - sin (x))

证明-cos(x)=sin(x+pi/2)

p ro v e - cos (x) = sin (x + \frac{π}{2})

证明 cos(4t)=1-8sin^2(t)cos^2(t)

p ro v e cos (4 t) = 1 - 8 sin^{2} (t) cos^{2} (t)

证明 sec(0)sin(0)=tan(0)

p ro v e sec (0) sin (0) = tan (0)