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 (cot(θ)-csc(θ))/(cot(θ)+csc(θ))=(1-2cos(θ)+cos^2(θ))/(-sin^2(θ))

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )}

用 sin, cos 表示

\frac{cot ( θ ) - csc ( θ )}{cot ( θ ) + csc ( θ )}

使用基本三角恒等式:

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

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - csc ( θ )}{\frac{c o s ( θ )}{s i n ( θ )} + csc ( θ )}

使用基本三角恒等式:

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

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}}

化简

\frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}} : \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

\frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )} + \frac{1}{s i n ( θ )}}

合并分式

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

使用法则

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

= \frac{cos ( θ ) + 1}{sin ( θ )}

= \frac{\frac{c o s ( θ )}{s i n ( θ )} - \frac{1}{s i n ( θ )}}{\frac{c o s ( θ ) + 1}{s i n ( θ )}}

合并分式

\frac{cos ( θ )}{sin ( θ )} - \frac{1}{sin ( θ )} : \frac{cos ( θ ) - 1}{sin ( θ )}

使用法则

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

= \frac{cos ( θ ) - 1}{sin ( θ )}

= \frac{\frac{c o s ( θ ) - 1}{s i n ( θ )}}{\frac{c o s ( θ ) + 1}{s i n ( θ )}}

分式相除:

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

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

约分:

sin (θ)

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

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

调整右侧

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

去除括号:

(a) = a

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

消掉

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

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

分解

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

1 - 2 cos (θ) + cos^{2} (θ)

改写成标准形式

a x^{2} + b x + c

= cos^{2} (θ) - 2 cos (θ) + 1

将

cos^{2} (θ) - 2 cos (θ) + 1

改写为

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

cos^{2} (θ) - 2 cos (θ) + 1

将

1

改写为

1^{2}

= cos^{2} (θ) - 2 cos (θ) + 1^{2}

将

2 cos (θ)

改写为

2 cos (θ) \cdot 1

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

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

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = cos (θ), b = 1

= (cos (θ) - 1)^{2}

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

分解

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

1 - cos^{2} (θ)

因式分解出通项

- 1

= - (cos^{2} (θ) - 1)

分解

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

cos^{2} (θ) - 1

将

1

改写为

1^{2}

= cos^{2} (θ) - 1^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

约分:

cos (θ) - 1

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

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

使用法则

- (- a) = a

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

= \frac{cos ( θ ) - 1}{cos ( θ ) + 1}

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 sin^2(θ/2)=(csc(θ)-cot(θ))/(2csc(θ))

p ro v e sin^{2} (\frac{θ}{2}) = \frac{csc ( θ ) - cot ( θ )}{2 csc ( θ )}

证明 cos(2x)=(cot^2(x)-1)/(cot^2(x)+1)

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

证明 sin^2(x)-2cos^2(x)=3sin^2(x)-2

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

证明 tan(a)=tan(a)csc^2(a)+cot(-a)

p ro v e tan (a) = tan (a) csc^{2} (a) + cot (- a)

证明 7csc^2(x)-5cot^2(x)=2csc^2(x)+5

p ro v e 7 csc^{2} (x) - 5 cot^{2} (x) = 2 csc^{2} (x) + 5