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受欢迎的三角函数 >

证明 cot(2u)=(cot^2(u)-1)/(2cot(u))

解答

证明

cot (2 u) = \frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

解答

真

求解步骤

cot (2 u) = \frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

调整右侧

\frac{cot ^{2} ( u ) - 1}{2 cot ( u )}

用 sin, cos 表示

\frac{- 1 + cot ^{2} ( u )}{2 cot ( u )}

使用基本三角恒等式:

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

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

化简

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

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

乘

2 \cdot \frac{cos ( u )}{sin ( u )} : \frac{2 cos ( u )}{sin ( u )}

2 \cdot \frac{cos ( u )}{sin ( u )}

分式相乘:

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

= \frac{cos ( u ) \cdot 2}{sin ( u )}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用分式法则:

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

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

化简

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

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

将项转换为分式:

1 = \frac{1 sin ^{2} ( u )}{sin ^{2} ( u )}

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

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 ^{2} ( u ) + cos ^{2} ( u )}{sin ^{2} ( u )}

乘以:

1 \cdot sin^{2} (u) = sin^{2} (u)

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

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

乘

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

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

分式相乘:

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

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

约分:

sin (u)

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

\frac{cos ^{2} ( u ) - sin ^{2} ( u )}{2 cos ( u ) sin ( u )}

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

使用倍角公式:

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

= \frac{cos ( 2 u )}{sin ( 2 u )}

= \frac{cos ( 2 u )}{sin ( 2 u )}

使用基本三角恒等式:

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

= cot (2 u)

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(2x))/2 =sin(x)cos(x)

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

证明 sin^2(z)+cos^2(z)=1

p ro v e sin^{2} (z) + cos^{2} (z) = 1

证明 1-cot(x)=sqrt(csc^2(x)-2cot(x))

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

证明 (csc^2(x))/(cot^2(x))=1+tan^2(x)

p ro v e \frac{csc ^{2} ( x )}{cot ^{2} ( x )} = 1 + tan^{2} (x)

证明 sin(x/2)=sqrt((1+cos(x))/2)

p ro v e sin (\frac{x}{2}) = \frac{1 + cos ( x )}{2}