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受欢迎的三角函数 >

证明 (cot(x)-csc(x))(cos(x)+1)=-sin(x)

解答

证明

(cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

解答

真

求解步骤

(cot (x) - csc (x)) (cos (x) + 1) = - sin (x)

调整左侧

(cot (x) - csc (x)) (cos (x) + 1)

用 sin, cos 表示

(1 + cos (x)) (cot (x) - csc (x))

使用基本三角恒等式:

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

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

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

化简

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

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

化简

\frac{cos ( x )}{sin ( x )} - \frac{1}{sin ( x )} : \frac{cos ( x ) - 1}{sin ( x )}

\frac{cos ( x )}{sin ( x )} - \frac{1}{sin ( x )}

使用法则

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

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

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

分式相乘:

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

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

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

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

使用三角恒等式改写

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

乘开

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

(cos (x) + 1) (cos (x) - 1)

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = cos (x), b = 1

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

使用法则

1^{a} = 1

1^{2} = 1

= cos^{2} (x) - 1

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

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

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

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

约分:

sin (x)

= - 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 csc^4(θ)-cot^4(θ)=2csc^2(θ)-1

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 csc^{2} (θ) - 1

证明 (cos(θ)cot(θ))/(1-sin(θ))-1=csc(θ)

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

证明 csc(θ)-sin(θ)=cos(θ)cot(θ)

p ro v e csc (θ) - sin (θ) = cos (θ) cot (θ)

证明 sec(x)(1-sin^2(x))=cos(x)

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

证明 (1-cos^2(θ))(1+cot^2(θ))=1

p ro v e (1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1