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受欢迎的三角函数 >

证明 (cos(θ))/(1-sin(θ))=(sin(θ)-csc(θ))/(cos(θ)-cot(θ))

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

化简

cos (θ) - \frac{cos ( θ )}{sin ( θ )} : \frac{cos ( θ ) sin ( θ ) - cos ( θ )}{sin ( θ )}

cos (θ) - \frac{cos ( θ )}{sin ( θ )}

将项转换为分式:

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

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

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

化简

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

- \frac{1}{sin ( θ )} + sin (θ)

将项转换为分式:

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

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

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

- 1 + 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)

= - 1 + sin^{2} (θ)

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

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

分式相除:

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

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

约分:

sin (θ)

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

分解

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

- 1 + sin^{2} (θ)

将

1

改写为

1^{2}

= sin^{2} (θ) - 1^{2}

使用平方差公式:

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

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

= (sin (θ) + 1) (sin (θ) - 1)

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

因式分解出通项

cos (θ)

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

约分:

sin (θ) - 1

= \frac{sin ( θ ) + 1}{cos ( θ )}

= \frac{sin ( θ ) + 1}{cos ( θ )}

= \frac{1 + sin ( θ )}{cos ( θ )}

使用三角恒等式改写

乘以

\frac{1 - sin ( θ )}{1 - sin ( θ )}

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

乘开

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

(1 + sin (θ)) (1 - sin (θ))

使用平方差公式:

(a + b) (a - b) = a^{2} - b^{2}

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

使用毕达哥拉斯恒等式:

1 = cos^{2} (θ) + sin^{2} (θ)

1 - sin^{2} (θ) = cos^{2} (θ)

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

约分:

cos (θ)

= \frac{cos ( θ )}{1 - sin ( θ )}

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

p ro v e \frac{cot ( θ ) sec ( θ )}{csc ( θ )} = 1

证明 sin^2(x)cos(x)sec(x)=1-cos^2(x)

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

证明 6sin(x)cos(x)=3sin(2x)

p ro v e 6 sin (x) cos (x) = 3 sin (2 x)

证明 (tan^2(θ))/(sec(θ))=sin(θ)tan(θ)

p ro v e \frac{tan ^{2} ( θ )}{sec ( θ )} = sin (θ) tan (θ)

证明 csc^2(x)(1-sin^2(x))=cot^2(x)

p ro v e csc^{2} (x) (1 - sin^{2} (x)) = cot^{2} (x)