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受欢迎的三角函数 >

证明 cot^2(t)-cos^2(t)=cot^2(t)cos^2(t)

解答

证明

cot^{2} (t) - cos^{2} (t) = cot^{2} (t) cos^{2} (t)

解答

真

求解步骤

cot^{2} (t) - cos^{2} (t) = cot^{2} (t) cos^{2} (t)

调整左侧

cot^{2} (t) - cos^{2} (t)

用 sin, cos 表示

- cos^{2} (t) + cot^{2} (t)

使用基本三角恒等式:

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

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

化简

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

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

使用指数法则:

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

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

将项转换为分式:

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

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

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

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

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

分解

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

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

分解

cos^{2} (t) - cos^{2} (t) sin^{2} (t) : cos^{2} (t) (1 - sin^{2} (t))

cos^{2} (t) - cos^{2} (t) sin^{2} (t)

因式分解出通项

cos^{2} (t)

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

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

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

使用三角恒等式改写

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

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} ( t ) cos ^{2} ( t )}{sin ^{2} ( t )}

cos^{2} (t) cos^{2} (t) = cos^{4} (t)

cos^{2} (t) cos^{2} (t)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (t) cos^{2} (t) = cos^{2 + 2} (t)

= cos^{2 + 2} (t)

数字相加:

2 + 2 = 4

= cos^{4} (t)

= \frac{cos ^{4} ( t )}{sin ^{2} ( t )}

= \frac{cos ^{4} ( t )}{sin ^{2} ( t )}

使用三角恒等式改写

= \frac{cos ^{3} ( t )}{sin ( t )} \cdot \frac{cos ( t )}{sin ( t )}

使用基本三角恒等式:

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

\frac{cos ^{3} ( t ) cot ( t )}{sin ( t )}

= cos^{2} (t) cot (t) \frac{cos ( t )}{sin ( t )}

使用基本三角恒等式:

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

cos^{2} (t) cot (t) cot (t)

化简

cos^{2} (t) cot (t) cot (t) : cos^{2} (t) cot^{2} (t)

cos^{2} (t) cot (t) cot (t)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cot (t) cot (t) = cot^{1 + 1} (t)

= cos^{2} (t) cot^{1 + 1} (t)

数字相加:

1 + 1 = 2

= cos^{2} (t) cot^{2} (t)

cos^{2} (t) cot^{2} (t)

cos^{2} (t) cot^{2} (t)

= cot^{2} (t) cos^{2} (t)

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 cot(θ)=(cos(θ))/(sin(θ))

p ro v e cot (θ) = \frac{cos ( θ )}{sin ( θ )}

证明 (1-cot(x))^2=csc^2(x)-2cot(x)

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

证明 1/(cot(x)(1-cos(2x)))=csc(2x)

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

证明 (tan(θ)+cot(θ))cos(θ)=csc(θ)

p ro v e (tan (θ) + cot (θ)) cos (θ) = csc (θ)

证明 cos^4(t)-sin^4(t)=1-2sin^2(t)

p ro v e cos^{4} (t) - sin^{4} (t) = 1 - 2 sin^{2} (t)