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受欢迎的三角函数 >

证明 (1+cos(θ))/(tan(θ)+sin(θ))=cot(θ)

解答

证明

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

解答

真

求解步骤

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

调整左侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

化简

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

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

将项转换为分式:

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

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

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

使用分式法则:

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

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

因式分解出通项

sin (θ)

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

约分:

1 + cos (θ)

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

= \frac{cos ( θ )}{sin ( θ )}

使用基本三角恒等式:

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

= cot (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^2(x)-1=(cos(x))/(tan(x)sin(x))

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

证明 (tan(x)sec(-x))cos(x)=sin(x)-1

p ro v e (tan (x) sec (- x)) cos (x) = sin (x) - 1

证明 (1-cos(x))*(1-cos(x))=(1-cos(x))^2

p ro v e (1 - cos (x)) \cdot (1 - cos (x)) = (1 - cos (x))^{2}

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p ro v e 2 cos^{2} (x) = 0

证明 (csc(-x))/(sec(-x))=-tan(pi/2-x)

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