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 sin^2(x)*(1+cot^2(x))=1

解答

证明

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

解答

真

求解步骤

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

调整左侧

sin^{2} (x) (1 + cot^{2} (x))

用 sin, cos 表示

(1 + cot^{2} (x)) sin^{2} (x)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

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

分式相乘:

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

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

约分:

sin^{2} (x)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

p ro v e cos (x + 1) = cos (- x - 1)

证明 cot^2(x)+1= 1/(sin^2(x))

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

证明 (sin(2A))/(1-cos(2A))=cot(A)

p ro v e \frac{sin ( 2 A )}{1 - cos ( 2 A )} = cot (A)

证明 sin(6x)=2sin(3x)cos(3x)

p ro v e sin (6 x) = 2 sin (3 x) cos (3 x)

证明 sin(4x)=4sin(x)cos(x)

p ro v e sin (4 x) = 4 sin (x) cos (x)