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受欢迎的三角函数 >

证明 (cos^2(A))/(cos(A)-sin(A))+(sin(A))/(1-cot(A))=sin(A)+cos(A)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}} : sin (A) + cos (A)

\frac{cos ^{2} ( A )}{cos ( A ) - sin ( A )} + \frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

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

\frac{sin ( A )}{1 - \frac{c o s ( A )}{s i n ( A )}}

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

= \frac{sin ( A )}{\frac{s i n ( A ) - c o s ( A )}{s i n ( A )}}

使用分式法则:

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

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

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

sin (A) sin (A)

使用指数法则:

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

sin (A) sin (A) = sin^{1 + 1} (A)

= sin^{1 + 1} (A)

数字相加:

1 + 1 = 2

= sin^{2} (A)

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

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

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

的最小公倍数:

- (cos (A) - sin (A))

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

最小公倍数 (LCM)

sin (A) - cos (A) = - (cos (A) - sin (A))

= cos (A) - sin (A), - (cos (A) - sin (A))

计算出由出现在

cos (A) - sin (A)

或

sin (A) - cos (A)

中的因子组成的表达式

= - (cos (A) - sin (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

- (cos (A) - sin (A))

对于

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

将分母和分子乘以

- 1

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

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

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

消掉

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

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

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

消掉

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

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

cos (A) - sin (A) = - (sin (A) - cos (A))

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

整理后得

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

约分:

sin (A) - cos (A)

= - (sin (A) + cos (A))

= - (sin (A) + cos (A))

取负

- (sin (A) + cos (A)) = - sin (A) - cos (A)

= - sin (A) - cos (A)

= - (- sin (A) - cos (A))

打开括号

= - (- sin (A)) - (- cos (A))

使用加减运算法则

- (- a) = a

= sin (A) + cos (A)

= sin (A) + cos (A)

= cos (A) + sin (A)

= sin (A) + cos (A)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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