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 tan(x)=tan(x)csc^2(x)+cot(-x)

解答

证明

tan (x) = tan (x) csc^{2} (x) + cot (- x)

解答

真

求解步骤

tan (x) = tan (x) csc^{2} (x) + cot (- x)

调整左侧

tan (x)

用 sin, cos 表示

tan (x)

使用基本三角恒等式:

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

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

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

调整右侧

tan (x) csc^{2} (x) + cot (- x)

使用负角恒等式:

cot (- x) = - cot (x)

= - cot (x) + csc^{2} (x) tan (x)

用 sin, cos 表示

- cot (x) + csc^{2} (x) tan (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

= \frac{sin ( x ) ( \frac{1}{s i n ( x )} ) ^{2}}{cos ( x )}

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

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{\frac{1}{s i n ^{2} ( x )} sin ( x )}{cos ( x )}

乘

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

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

分式相乘:

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

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

乘以:

1 \cdot sin (x) = sin (x)

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

约分:

sin (x)

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

= \frac{\frac{1}{s i n ( x )}}{cos ( x )}

使用分式法则:

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

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

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

sin (x), sin (x) cos (x)

的最小公倍数:

sin (x) cos (x)

sin (x), sin (x) cos (x)

最小公倍数 (LCM)

计算出由出现在

sin (x)

或

sin (x) cos (x)

中的因子组成的表达式

= sin (x) cos (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (x) cos (x)

对于

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

将分母和分子乘以

cos (x)

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

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

因为分母相等，所以合并分式:

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

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

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

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

使用三角恒等式改写

- cot (x) + csc^{2} (x) tan (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)

= - cot (x) + csc^{2} (x) tan (x)

= - cot (x) + csc^{2} (x) tan (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 tan(8x)=(2tan(x))/(1-tan^2(x))

p ro v e tan (8 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

证明 (cos(x)sin(x))/(sin(x)cos(x))=1

p ro v e \frac{cos ( x ) sin ( x )}{sin ( x ) cos ( x )} = 1

证明 1/(cot^2(x))+sec(x)cos(x)=1

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

证明 3-3(cos(x)-sin(x))^2=-3cos(2x)

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

证明 csc^2(x)+tan^2(x)-1=tan^2(x)

p ro v e csc^{2} (x) + tan^{2} (x) - 1 = tan^{2} (x)