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受欢迎的三角函数 >

证明 tan^2(x)=((1-cos(2x)))/(1+cos(2x))

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

数字相加:

1 + 1 = 2

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

乘开

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

分解

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

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

改写为

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

因式分解出通项

2

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

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

数字相除:

\frac{2}{2} = 1

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

去除括号:

(- a) = - a

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

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

使用基本三角恒等式:

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

tan (x) tan (x)

化简

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

tan (x) tan (x)

使用指数法则:

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

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

= tan^{1 + 1} (x)

数字相加:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (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 cos(0)tan(0)=sin(0)

p ro v e cos (0) tan (0) = sin (0)

证明 cos(pi/6)= 4/3 cos^3(pi/6)

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

证明 tan(-pi/4)=-1

p ro v e tan (- \frac{π}{4}) = - 1

证明 sin^2(x)= 1/(1+cot^2(x))

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

证明 (2sec^2(x)-2tan^2(x))/(csc(x))=sin(2x)sec(x)

p ro v e \frac{2 sec ^{2} ( x ) - 2 tan ^{2} ( x )}{csc ( x )} = sin (2 x) sec (x)