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受欢迎的三角函数 >

证明 tan(θ)csc^2(θ)-tan(θ)=cot(θ)

解答

证明

tan (θ) csc^{2} (θ) - tan (θ) = cot (θ)

解答

真

求解步骤

tan (θ) csc^{2} (θ) - tan (θ) = cot (θ)

调整左侧

tan (θ) csc^{2} (θ) - tan (θ)

用 sin, cos 表示

- tan (θ) + csc^{2} (θ) tan (θ)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

csc (x) = \frac{1}{sin ( x )}

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

化简

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

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

(\frac{1}{sin ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )} = \frac{1}{sin ( θ ) cos ( θ )}

(\frac{1}{sin ( θ )})^{2} \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{sin ( θ ) ( \frac{1}{s i n ( θ )} ) ^{2}}{cos ( θ )}

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

(\frac{1}{sin ( θ )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

= \frac{\frac{1}{s i n ^{2} ( θ )} sin ( θ )}{cos ( θ )}

乘

sin (θ) \frac{1}{sin ^{2} ( θ )} : \frac{1}{sin ( θ )}

sin (θ) \frac{1}{sin ^{2} ( θ )}

分式相乘:

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

= \frac{1 \cdot sin ( θ )}{sin ^{2} ( θ )}

乘以:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{sin ^{2} ( θ )}

约分:

sin (θ)

= \frac{1}{sin ( θ )}

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

使用分式法则:

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

= \frac{1}{sin ( θ ) cos ( θ )}

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

cos (θ), sin (θ) cos (θ)

的最小公倍数:

cos (θ) sin (θ)

cos (θ), sin (θ) cos (θ)

最小公倍数 (LCM)

计算出由出现在

cos (θ)

或

sin (θ) cos (θ)

中的因子组成的表达式

= cos (θ) sin (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

cos (θ) sin (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 )}{cos ( θ ) sin ( θ )}

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

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

= \frac{1 - sin ^{2} ( θ )}{cos ( θ ) sin ( θ )}

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

约分:

cos (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

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p ro v e cos (α + β) + cos (α - β) = 2 cos (α) cos (β)

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

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

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p ro v e 9 sec^{2} (θ) - 5 tan^{2} (θ) = 5 + 4 sec^{2} (θ)

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