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受欢迎的三角函数 >

证明 (1+csc(θ))/(sec(θ))-cot(θ)=cos(θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

\frac{1 + csc ( θ )}{sec ( θ )} - cot (θ)

用 sin, cos 表示

- cot (θ) + \frac{1 + csc ( θ )}{sec ( θ )}

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用分式法则:

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

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

化简

1 + \frac{1}{sin ( θ )} : \frac{sin ( θ ) + 1}{sin ( θ )}

1 + \frac{1}{sin ( θ )}

将项转换为分式:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

乘以:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

使用分式法则:

\frac{a}{1} = a

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

分式相乘:

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

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

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

使用法则

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

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

分解

- cos (θ) + (sin (θ) + 1) cos (θ) : cos (θ) sin (θ)

- cos (θ) + (sin (θ) + 1) cos (θ)

因式分解出通项

cos (θ)

= cos (θ) (- 1 + 1 + sin (θ))

整理后得

= cos (θ) sin (θ)

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

约分:

sin (θ)

= cos (θ)

= cos (θ)

= cos (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 sin(2θ)=(2cot(θ))/(1+cot^2(θ))

p ro v e sin (2 θ) = \frac{2 cot ( θ )}{1 + cot ^{2} ( θ )}

证明 tan(θ)=(1-cos(2θ))/(sin(2θ))

p ro v e tan (θ) = \frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

证明 (sin^2(x)cot(x))/(cos(x))=sin(x)

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

证明 csc(2θ)+cot(2θ)=cot(θ)

p ro v e csc (2 θ) + cot (2 θ) = cot (θ)

证明-tan(x)cos(x)=sin(-x)

p ro v e - tan (x) cos (x) = sin (- x)