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(x))/(sin(x)-1)=-sin(x)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

消掉

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

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

分解

1 - sin^{2} (x) : - (sin (x) + 1) (sin (x) - 1)

1 - sin^{2} (x)

因式分解出通项

- 1

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

分解

sin^{2} (x) - 1 : (sin (x) + 1) (sin (x) - 1)

sin^{2} (x) - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (sin (x) + 1) (sin (x) - 1)

= - (sin (x) + 1) (sin (x) - 1)

= - \frac{( sin ( x ) + 1 ) ( sin ( x ) - 1 )}{sin ( x ) - 1}

约分:

sin (x) - 1

= - (sin (x) + 1)

取负

- (sin (x) + 1) = - sin (x) - 1

= - sin (x) - 1

= 1 + (- sin (x) - 1)

去除括号:

(- a) = - a

= 1 - sin (x) - 1

对同类项分组

= - sin (x) + 1 - 1

1 - 1 = 0

= - sin (x)

= - sin (x)

= - sin (x)

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(3x)=3cos^2(x)sin(x)-sin^3(x)

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

证明 cos(3x)+cos(x)=4cos^3(x)-2cos(x)

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

证明 (cos(x))/(sin(x))=cot(x)

p ro v e \frac{cos ( x )}{sin ( x )} = cot (x)

证明 (cot(2θ))/(cos(2θ))=csc(2θ)

p ro v e \frac{cot ( 2 θ )}{cos ( 2 θ )} = csc (2 θ)

证明 cos^3(θ)+cos(θ)sin^2(θ)=cos(θ)

p ro v e cos^{3} (θ) + cos (θ) sin^{2} (θ) = cos (θ)