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受欢迎的三角函数 >

证明 tan(A)=tan(A)csc^2(A)+cot(-A)

解答

证明

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

解答

真

求解步骤

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

调整左侧

tan (A)

用 sin, cos 表示

tan (A)

使用基本三角恒等式:

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

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

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

调整右侧

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

使用负角恒等式:

cot (- x) = - cot (x)

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

乘

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

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

分式相乘:

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

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

乘以:

1 \cdot sin (A) = sin (A)

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

约分:

sin (A)

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

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

使用分式法则:

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

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

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

sin (A), sin (A) cos (A)

的最小公倍数:

sin (A) cos (A)

sin (A), sin (A) cos (A)

最小公倍数 (LCM)

计算出由出现在

sin (A)

或

sin (A) cos (A)

中的因子组成的表达式

= sin (A) cos (A)

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 (A) cos (A)

对于

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

将分母和分子乘以

cos (A)

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

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

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} ( A ) + 1}{sin ( A ) cos ( A )}

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

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

使用三角恒等式改写

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

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 (A) + csc^{2} (A) tan (A)

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

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 1/(1+sin(x))=(sec(x)-tan(x))sec(x)

p ro v e \frac{1}{1 + sin ( x )} = (sec (x) - tan (x)) sec (x)

证明 tan(a)+cot(a)= 2/(sin(2a))

p ro v e tan (a) + cot (a) = \frac{2}{sin ( 2 a )}

证明 (1+sin(α))/(cos(α))=(cos(α))/(1-sin(α))

p ro v e \frac{1 + sin ( α )}{cos ( α )} = \frac{cos ( α )}{1 - sin ( α )}

证明 2csc^2(y)= 1/(1-cos(y))+1/(1+cos(y))

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

证明 cot(2x)=(cos(2x))/(sin(2x))

p ro v e cot (2 x) = \frac{cos ( 2 x )}{sin ( 2 x )}