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人気のある三角関数 >

証明する cos(3A)=4cos^3(A)-3cos(A)

解

証明する

cos (3 A) = 4 cos^{3} (A) - 3 cos (A)

解

真

解答ステップ

cos (3 A) = 4 cos^{3} (A) - 3 cos (A)

左側を操作する

cos (3 A)

三角関数の公式を使用して書き換える

cos (3 A)

書き換え

= cos (2 A + A)

角の和の公式を使用する:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 A) cos (A) - sin (2 A) sin (A)

2倍角の公式を使用:

sin (2 A) = 2 sin (A) cos (A)

= cos (2 A) cos (A) - 2 sin (A) cos (A) sin (A)

簡素化

cos (2 A) cos (A) - 2 sin (A) cos (A) sin (A) : cos (A) cos (2 A) - 2 sin^{2} (A) cos (A)

cos (2 A) cos (A) - 2 sin (A) cos (A) sin (A)

2 sin (A) cos (A) sin (A) = 2 sin^{2} (A) cos (A)

2 sin (A) cos (A) sin (A)

指数の規則を適用する:

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

sin (A) sin (A) = sin^{1 + 1} (A)

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

数を足す:

1 + 1 = 2

= 2 cos (A) sin^{2} (A)

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

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

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

2倍角の公式を使用:

cos (2 A) = 2 cos^{2} (A) - 1

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

ピタゴラスの公式を使用する:

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

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

= (2 cos^{2} (A) - 1) cos (A) - 2 (1 - cos^{2} (A)) cos (A)

拡張

(2 cos^{2} (A) - 1) cos (A) - 2 (1 - cos^{2} (A)) cos (A) : 4 cos^{3} (A) - 3 cos (A)

(2 cos^{2} (A) - 1) cos (A) - 2 (1 - cos^{2} (A)) cos (A)

= cos (A) (2 cos^{2} (A) - 1) - 2 cos (A) (1 - cos^{2} (A))

拡張

cos (A) (2 cos^{2} (A) - 1) : 2 cos^{3} (A) - cos (A)

cos (A) (2 cos^{2} (A) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = cos (A), b = 2 cos^{2} (A), c = 1

= cos (A) 2 cos^{2} (A) - cos (A) 1

= 2 cos^{2} (A) cos (A) - 1 cos (A)

簡素化

2 cos^{2} (A) cos (A) - 1 \cdot cos (A) : 2 cos^{3} (A) - cos (A)

2 cos^{2} (A) cos (A) - 1 cos (A)

2 cos^{2} (A) cos (A) = 2 cos^{3} (A)

2 cos^{2} (A) cos (A)

指数の規則を適用する:

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

cos^{2} (A) cos (A) = cos^{2 + 1} (A)

= 2 cos^{2 + 1} (A)

数を足す:

2 + 1 = 3

= 2 cos^{3} (A)

1 \cdot cos (A) = cos (A)

1 cos (A)

乗算:

1 \cdot cos (A) = cos (A)

= cos (A)

= 2 cos^{3} (A) - cos (A)

= 2 cos^{3} (A) - cos (A)

= 2 cos^{3} (A) - cos (A) - 2 (1 - cos^{2} (A)) cos (A)

拡張

- 2 cos (A) (1 - cos^{2} (A)) : - 2 cos (A) + 2 cos^{3} (A)

- 2 cos (A) (1 - cos^{2} (A))

分配法則を適用する:

a (b - c) = ab - a c

a = - 2 cos (A), b = 1, c = cos^{2} (A)

= - 2 cos (A) 1 - (- 2 cos (A)) cos^{2} (A)

マイナス・プラスの規則を適用する

- (- a) = a

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

簡素化

- 2 \cdot 1 \cdot cos (A) + 2 cos^{2} (A) cos (A) : - 2 cos (A) + 2 cos^{3} (A)

- 2 \cdot 1 cos (A) + 2 cos^{2} (A) cos (A)

2 \cdot 1 \cdot cos (A) = 2 cos (A)

2 \cdot 1 cos (A)

数を乗じる:

2 \cdot 1 = 2

= 2 cos (A)

2 cos^{2} (A) cos (A) = 2 cos^{3} (A)

2 cos^{2} (A) cos (A)

指数の規則を適用する:

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

cos^{2} (A) cos (A) = cos^{2 + 1} (A)

= 2 cos^{2 + 1} (A)

数を足す:

2 + 1 = 3

= 2 cos^{3} (A)

= - 2 cos (A) + 2 cos^{3} (A)

= - 2 cos (A) + 2 cos^{3} (A)

= 2 cos^{3} (A) - cos (A) - 2 cos (A) + 2 cos^{3} (A)

簡素化

2 cos^{3} (A) - cos (A) - 2 cos (A) + 2 cos^{3} (A) : 4 cos^{3} (A) - 3 cos (A)

2 cos^{3} (A) - cos (A) - 2 cos (A) + 2 cos^{3} (A)

条件のようなグループ

= 2 cos^{3} (A) + 2 cos^{3} (A) - cos (A) - 2 cos (A)

類似した元を足す:

2 cos^{3} (A) + 2 cos^{3} (A) = 4 cos^{3} (A)

= 4 cos^{3} (A) - cos (A) - 2 cos (A)

類似した元を足す:

- cos (A) - 2 cos (A) = - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (A)

= 4 cos^{3} (A) - 3 cos (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(3θ)=cos^3(θ)-3sin^2(θ)cos(θ)

p ro v e cos (3 θ) = cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

証明する cos^2(B)-sin^2(B)=2cos^2(B)-1

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

証明する csc^2(x)cos^2(x)= 1/(tan^2(x))

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

証明する 1+tan^2(t)=sec^2(t)

p ro v e 1 + tan^{2} (t) = sec^{2} (t)

証明する cos(4x)=2cos^2(2x)-1

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