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受欢迎的三角函数 >

证明 (cot(θ)-tan(θ))/(cot(θ)+tan(θ))=cos(2θ)

解答

证明

\frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )} = cos (2 θ)

解答

真

求解步骤

\frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )} = cos (2 θ)

调整左侧

\frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )}

用 sin, cos 表示

\frac{cot ( θ ) - tan ( θ )}{cot ( θ ) + tan ( θ )}

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

化简

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

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

sin (θ), cos (θ)

的最小公倍数:

sin (θ) cos (θ)

sin (θ), cos (θ)

最小公倍数 (LCM)

计算出由出现在

sin (θ)

或

cos (θ)

中的因子组成的表达式

= sin (θ) cos (θ)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (θ) cos (θ)

对于

\frac{cos ( θ )}{sin ( θ )} :

将分母和分子乘以

cos (θ)

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

对于

\frac{sin ( θ )}{cos ( θ )} :

将分母和分子乘以

sin (θ)

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

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

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

化简

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

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

sin (θ), cos (θ)

的最小公倍数:

sin (θ) cos (θ)

sin (θ), cos (θ)

最小公倍数 (LCM)

计算出由出现在

sin (θ)

或

cos (θ)

中的因子组成的表达式

= sin (θ) cos (θ)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

sin (θ) cos (θ)

对于

\frac{cos ( θ )}{sin ( θ )} :

将分母和分子乘以

cos (θ)

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

对于

\frac{sin ( θ )}{cos ( θ )} :

将分母和分子乘以

sin (θ)

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

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

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

分式相除:

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

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

约分:

sin (θ)

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

约分:

cos (θ)

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用毕达哥拉斯恒等式:

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

= \frac{cos ( 2 θ )}{1}

使用法则

\frac{a}{1} = a

= cos (2 θ)

= cos (2 θ)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

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