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+tan^2(u))(1-sin^2(u))=1

解答

证明

(1 + tan^{2} (u)) (1 - sin^{2} (u)) = 1

解答

真

求解步骤

(1 + tan^{2} (u)) (1 - sin^{2} (u)) = 1

调整左侧

(1 + tan^{2} (u)) (1 - sin^{2} (u))

使用三角恒等式改写

(1 + tan^{2} (u)) (1 - sin^{2} (u))

使用毕达哥拉斯恒等式:

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

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

= (1 + tan^{2} (u)) cos^{2} (u)

= (1 + tan^{2} (u)) cos^{2} (u)

用 sin, cos 表示

(1 + tan^{2} (u)) cos^{2} (u)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

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

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

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

分式相乘:

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

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

约分:

cos^{2} (u)

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

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

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

使用三角恒等式改写

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

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 sec^3(x)= 1/(cos^3(x))

p ro v e sec^{3} (x) = \frac{1}{cos ^{3} ( x )}

证明 4cos^2(B)-4sin^2(B)=4-8sin^2(B)

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

证明 cos(4A)=8cos^4(A)-8cos^2(A)+1

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

证明 cos(4B)=1-8sin^2(B)+8sin^4(B)

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

证明 sin(pi/2-u)=cos(u)

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