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受欢迎的三角函数 >

证明 cot(2β)=(cot^2(β)-1)/(2cot(β))

解答

证明

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

解答

真

求解步骤

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

调整右侧

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

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

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

乘

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

2 \cdot \frac{cos ( β )}{sin ( β )}

分式相乘:

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

= \frac{cos ( β ) \cdot 2}{sin ( β )}

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

使用指数法则:

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

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

使用分式法则:

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

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

化简

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

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

将项转换为分式:

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

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

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

乘以:

1 \cdot sin^{2} (β) = sin^{2} (β)

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

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

乘

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

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

分式相乘:

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

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

约分:

sin (β)

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

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

使用分式法则:

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

使用倍角公式:

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

= \frac{cos ( 2 β )}{sin ( 2 β )}

= \frac{cos ( 2 β )}{sin ( 2 β )}

使用基本三角恒等式:

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

= cot (2 β)

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(x)-cot(x))/(csc(x))=1-cos(x)

p ro v e \frac{csc ( x ) - cot ( x )}{csc ( x )} = 1 - cos (x)

证明 arcsec(x)=cos(x)

p ro v e arcsec (x) = cos (x)

证明 (csc(θ)-cot(θ))(1+sec(θ))=tan(θ)

p ro v e (csc (θ) - cot (θ)) (1 + sec (θ)) = tan (θ)

证明 1/(sin(x)cos(x))=sec(x)csc(x)

p ro v e \frac{1}{sin ( x ) cos ( x )} = sec (x) csc (x)

证明 cot^2(x)+csc^2(x)=1+2cot^2(x)

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