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人気のある三角関数 >

証明する (1-cos^2(θ))(1+cot^2(θ))=1

解

証明する

(1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1

解

真

解答ステップ

(1 - cos^{2} (θ)) (1 + cot^{2} (θ)) = 1

左側を操作する

(1 - cos^{2} (θ)) (1 + cot^{2} (θ))

サイン, コサインで表わす

(1 + cot^{2} (θ)) (1 - cos^{2} (θ))

基本的な三角関数の公式を使用する:

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

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

簡素化

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

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

指数の規則を適用する:

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

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

結合

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

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

元を分数に変換する:

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

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

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

乗算:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

分数を乗じる:

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

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

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

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

(1 - cos^{2} (θ)) (cos^{2} (θ) + 1 - cos^{2} (θ))

簡素化

cos^{2} (θ) + 1 - cos^{2} (θ) : 1

cos^{2} (θ) + 1 - cos^{2} (θ)

条件のようなグループ

= cos^{2} (θ) - cos^{2} (θ) + 1

類似した元を足す:

cos^{2} (θ) - cos^{2} (θ) = 0

= 1

= 1 \cdot (- cos^{2} (θ) + 1)

乗算:

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

= (1 - cos^{2} (θ))

括弧を削除する:

(a) = a

= 1 - cos^{2} (θ)

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

規則を適用

\frac{a}{a} = 1

= 1

= 1

= 1

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 cot(x)sin(x)=cos(x)

p ro v e cot (x) sin (x) = cos (x)

証明する sin(2x)+cos(2x)-1=2sin(x)(cos(x)-sin(x))

p ro v e sin (2 x) + cos (2 x) - 1 = 2 sin (x) (cos (x) - sin (x))

証明する (csc^2(x)-1)sin(x)=cot(x)cos(x)

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

証明する cot(x)(1-cos^2(x))= 1/2 sin(2x)

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

証明する (sec^2(t))/(tan(t))=sec(t)csc(t)

p ro v e \frac{sec ^{2} ( t )}{tan ( t )} = sec (t) csc (t)