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受欢迎的三角函数 >

证明 (1-cos(a))/(1+cos(a))=tan^2(a/2)

解答

证明

\frac{1 - cos ( a )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

解答

真

求解步骤

\frac{1 - cos ( a )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

令

: u = \frac{a}{2}

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

证明

\frac{1 - cos ( 2 u )}{1 + cos ( 2 u )} = tan^{2} (u) :

真

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

调整左侧

\frac{1 - cos ( 2 u )}{1 + cos ( 2 u )}

使用三角恒等式改写

\frac{1 - cos ( 2 u )}{1 + cos ( 2 u )}

使用倍角公式:

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

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

化简

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

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

数字相加:

1 + 1 = 2

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

乘开

1 - (1 - 2 sin^{2} (u)) : 2 sin^{2} (u)

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

1 - 1 = 0

= 2 sin^{2} (u)

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

分解

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

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

改写为

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

因式分解出通项

2

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

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

数字相除:

\frac{2}{2} = 1

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

去除括号:

(- a) = - a

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

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

使用毕达哥拉斯恒等式:

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

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

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

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

使用三角恒等式改写

= \frac{sin ( u )}{cos ( u )} \cdot \frac{sin ( u )}{cos ( u )}

使用基本三角恒等式:

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

\frac{sin ( u ) tan ( u )}{cos ( u )}

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

使用基本三角恒等式:

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

tan (u) tan (u)

化简

tan (u) tan (u) : tan^{2} (u)

tan (u) tan (u)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{1 + 1} (u)

数字相加:

1 + 1 = 2

= tan^{2} (u)

tan^{2} (u)

tan^{2} (u)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

因此

\frac{1 - cos ( a )}{1 + cos ( a )} = tan^{2} (\frac{a}{2})

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^2(x)-cos(x)sec(x)=cot^2(x)

p ro v e csc^{2} (x) - cos (x) sec (x) = cot^{2} (x)

证明 (csc(x)+1)/(cot(x)+cos(x))=sec(x)

p ro v e \frac{csc ( x ) + 1}{cot ( x ) + cos ( x )} = sec (x)

证明 csc^2(θ/2)= 2/(1-cos(θ))

p ro v e csc^{2} (\frac{θ}{2}) = \frac{2}{1 - cos ( θ )}

证明 sec(x)cos(x)+tan^2(x)=sec^2(x)

p ro v e sec (x) cos (x) + tan^{2} (x) = sec^{2} (x)

证明 cot^2(θ)+1= 1/(sin^2(θ))

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