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受欢迎的三角函数 >

证明 cot(2b)=(cot^2(b)-1)/(2cot(b))

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

乘

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

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

分式相乘:

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

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

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

使用指数法则:

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

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

使用分式法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

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

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

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

乘

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

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

分式相乘:

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

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

约分:

sin (b)

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用倍角公式:

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

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

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

使用基本三角恒等式:

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

= cot (2 b)

我们已展示，在两侧可以有相同的形式

\Rightarrow 真

流行的例子

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p ro v e 2 cos (x) cos (y) = cos (x + y) + cos (x - y)

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

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p ro v e (1 + tan^{2} (u)) (1 - sin^{2} (u)) = 1