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人気のある三角関数 >

証明する cos^2(a)cot^2(a)=cot^2(a)-cos^2(a)

解

証明する

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

解

真

解答ステップ

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

右側を操作する

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

サイン, コサインで表わす

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

指数の規則を適用する:

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

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

元を分数に変換する:

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

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

分母が等しいので, 分数を組み合わせる:

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

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

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

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

因数

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

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

因数

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

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

共通項をくくり出す

cos^{2} (a)

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

cos^{2} (a) cos^{2} (a)

指数の規則を適用する:

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

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

= cos^{2 + 2} (a)

数を足す:

2 + 2 = 4

= cos^{4} (a)

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

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

三角関数の公式を使用して書き換える

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

基本的な三角関数の公式を使用する:

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

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

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

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する cos(θ)= 1/2

p ro v e cos (θ) = \frac{1}{2}

証明する csc(2x)= 1/(sin^2(x)-1)

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

証明する (1+tan^2(α))(1-sin^2(α))=1

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

証明する csc(θ)-cot(θ)=(1-cos(θ))/(sin(θ))

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

証明する tan(x)=tan(x)+2sin^2(x)

p ro v e tan (x) = tan (x) + 2 sin^{2} (x)