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受欢迎的三角函数 >

证明 (cos(θ)cot(θ))/(1-sin(θ))-1=csc(θ)

解答

证明

\frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

解答

真

求解步骤

\frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1 = csc (θ)

调整左侧

\frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )} - 1

用 sin, cos 表示

- 1 + \frac{cos ( θ ) cot ( θ )}{1 - sin ( θ )}

使用基本三角恒等式:

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

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

化简

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

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

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

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

乘

cos (θ) \frac{cos ( θ )}{sin ( θ )} : \frac{cos ^{2} ( θ )}{sin ( θ )}

cos (θ) \frac{cos ( θ )}{sin ( θ )}

分式相乘:

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

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

使用分式法则:

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

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

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

将项转换为分式:

1 = \frac{1 sin ( θ ) ( 1 - sin ( θ ) )}{sin ( θ ) ( 1 - sin ( θ ) )}

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

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 ( θ ) ( 1 - sin ( θ ) ) + cos ^{2} ( θ )}{sin ( θ ) ( 1 - sin ( θ ) )}

乘以:

1 \cdot sin (θ) = sin (θ)

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

\frac{1 - sin ^{2} ( θ ) - ( 1 - sin ( θ ) ) sin ( θ )}{( 1 - sin ( θ ) ) sin ( θ )}

乘开

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

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

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

乘开

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

- sin (θ) (1 - sin (θ))

使用分配律:

a (b - c) = ab - a c

a = - sin (θ), b = 1, c = sin (θ)

= - sin (θ) \cdot 1 - (- sin (θ)) sin (θ)

使用加减运算法则

- (- a) = a

= - 1 \cdot sin (θ) + sin (θ) sin (θ)

化简

- 1 \cdot sin (θ) + sin (θ) sin (θ) : - sin (θ) + sin^{2} (θ)

- 1 \cdot sin (θ) + sin (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

乘以:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

使用指数法则:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= sin^{2} (θ)

= - sin (θ) + sin^{2} (θ)

= - sin (θ) + sin^{2} (θ)

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

化简

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

1 - sin^{2} (θ) - sin (θ) + sin^{2} (θ)

对同类项分组

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

同类项相加:

- sin^{2} (θ) + sin^{2} (θ) = 0

= - sin (θ) + 1

= - sin (θ) + 1

= \frac{- sin ( θ ) + 1}{sin ( θ ) ( - sin ( θ ) + 1 )}

约分:

- sin (θ) + 1

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

使用三角恒等式改写

使用基本三角恒等式:

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

\frac{1}{\frac{1}{c s c ( θ )}}

化简

\frac{1}{\frac{1}{c s c ( θ )}}

使用分式法则:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( θ )}{1}

使用法则

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

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 csc(θ)-sin(θ)=cos(θ)cot(θ)

p ro v e csc (θ) - sin (θ) = cos (θ) cot (θ)

证明 sec(x)(1-sin^2(x))=cos(x)

p ro v e sec (x) (1 - sin^{2} (x)) = cos (x)

证明 (1-cos^2(θ))(1+cot^2(θ))=1

p ro v e (1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1

证明 cot(x)sin(x)=cos(x)

p ro v e cot (x) sin (x) = cos (x)

证明 sin(2x)+cos(2x)-1=2sin(x)(cos(x)-sin(x))

p ro v e sin (2 x) + cos (2 x) - 1 = 2 sin (x) (cos (x) - sin (x))