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-cos^2(A))(1+cot^2(A))=1

解答

证明

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

解答

真

求解步骤

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

调整左侧

(1 - cos^{2} (A)) (1 + cot^{2} (A))

用 sin, cos 表示

(1 + cot^{2} (A)) (1 - cos^{2} (A))

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 sin ^{2} ( A )}{sin ^{2} ( A )}

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

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 sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )}

乘以:

1 \cdot sin^{2} (A) = sin^{2} (A)

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

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

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 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + 1 - cos ^{2} ( A ) )}{1 - cos ^{2} ( A )}

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

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

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

(1 - cos^{2} (A)) (cos^{2} (A) + 1 - cos^{2} (A))

化简

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

cos^{2} (A) + 1 - cos^{2} (A)

对同类项分组

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

同类项相加:

cos^{2} (A) - cos^{2} (A) = 0

= 1

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

乘以:

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

= (1 - cos^{2} (A))

去除括号:

(a) = a

= 1 - cos^{2} (A)

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

使用法则

\frac{a}{a} = 1

= 1

= 1

= 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 4cos^2(x)+cos(2x)=5-6sin^2(x)

p ro v e 4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)

证明 (sin^2(2x))/4 =sin^2(x)cos^2(x)

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

证明 (sec(θ))/(tan(θ)+cot(θ))=sin(θ)

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

证明 (sin(2x))/(cos(2x)+sin^2(x))=2tan(x)

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

证明 7cos^2(x)+5sin^2(x)=6+cos(2x)

p ro v e 7 cos^{2} (x) + 5 sin^{2} (x) = 6 + cos (2 x)