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人気のある三角関数 >

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

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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)

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

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

証明する tan(x+pi/3)=(tan(x)+sqrt(3))/(1-sqrt(3)tan(x))

p ro v e tan (x + \frac{π}{3}) = \frac{tan ( x ) + 3}{1 - 3 tan ( x )}

証明する (1-2sin^2(x))/(1-sin^2(x))=1-tan(2x)

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

証明する sin(x+y)-sin(x-y)=2sin(y)cos(x)

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

証明する sin(2a)=2cos(a)sin(a)

p ro v e sin (2 a) = 2 cos (a) sin (a)