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受欢迎的三角函数 >

证明 (1-cos(2a))/(1+cos(2a))=tan^2(a)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

数字相加:

1 + 1 = 2

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

乘开

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

1 - 1 = 0

= 2 sin^{2} (a)

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

分解

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

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

改写为

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

因式分解出通项

2

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

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

数字相除:

\frac{2}{2} = 1

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

去除括号:

(- a) = - a

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

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

使用基本三角恒等式:

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

tan (a) tan (a)

化简

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

tan (a) tan (a)

使用指数法则:

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

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

= tan^{1 + 1} (a)

数字相加:

1 + 1 = 2

= tan^{2} (a)

tan^{2} (a)

tan^{2} (a)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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