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^2(θ)+1= 1/(sin^2(θ))

解答

证明

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

解答

真

求解步骤

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

调整左侧

cot^{2} (θ) + 1

用 sin, cos 表示

1 + cot^{2} (θ)

使用基本三角恒等式:

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

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

化简

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

1 + (\frac{cos ( θ )}{sin ( θ )})^{2}

使用指数法则:

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( 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 3cos(2x+pi/2)-2=6cos^2(x+pi/4)

p ro v e 3 cos (2 x + \frac{π}{2}) - 2 = 6 cos^{2} (x + \frac{π}{4})

证明 cos^4(b)-sin^4(b)=cos(2b)

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

证明 sin^2(x)sec^2(x)=sec^2(x)-1

p ro v e sin^{2} (x) sec^{2} (x) = sec^{2} (x) - 1

证明 (6cos^2(x)-6sin^2(x))=6-12sin^2(x)

p ro v e (6 cos^{2} (x) - 6 sin^{2} (x)) = 6 - 12 sin^{2} (x)

证明 1+1/(tan^2(x))=csc^2(x)

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