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受欢迎的三角函数 >

证明 (csc^2(x)-1)sec(x)=cot(x)csc(x)

解答

证明

(csc^{2} (x) - 1) sec (x) = cot (x) csc (x)

解答

真

求解步骤

(csc^{2} (x) - 1) sec (x) = cot (x) csc (x)

调整左侧

(csc^{2} (x) - 1) sec (x)

用 sin, cos 表示

(- 1 + csc^{2} (x)) sec (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

分式相乘:

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

= \frac{1 \cdot ( - 1 + ( \frac{1}{s i n ( x )} ) ^{2} )}{cos ( x )}

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

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

乘以:

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

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

去除括号:

(- a) = - a

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

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

化简

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

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

将项转换为分式:

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

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

乘以:

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

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

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{cos ^{2} ( x )}{cos ( x ) sin ^{2} ( x )}

约分:

cos (x)

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

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

调整右侧

cot (x) csc (x)

用 sin, cos 表示

cot (x) csc (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

分式相乘:

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

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

乘以:

cos (x) \cdot 1 = cos (x)

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

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

sin (x) sin (x)

使用指数法则:

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

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

= sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= sin^{2} (x)

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

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

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

p ro v e cot^{2} (y) (sec^{2} (y) - 1) = 1

证明 tan(x-pi/2)=-cot(x)

p ro v e tan (x - \frac{π}{2}) = - cot (x)

证明 sin(pi-θ)=sin(θ)

p ro v e sin (π - θ) = sin (θ)

证明 tan(x/2)=csc(x)-cot(x)

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

证明 tan(x)cot(x)=1

p ro v e tan (x) cot (x) = 1