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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 2cos(x)sin(y)=sin(x+y)-sin(x-y)

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

証明する (tan(2x)+cot(2x))/(csc(2x))=sec(2x)

p ro v e \frac{tan ( 2 x ) + cot ( 2 x )}{csc ( 2 x )} = sec (2 x)

証明する sin(3x)=sin(x)(4cos^2(x)-1)

p ro v e sin (3 x) = sin (x) (4 cos^{2} (x) - 1)

証明する sin^2(z)=(1-cos(2z))/2

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

証明する sin^2(pi/2-x)+sin^2(x)=1

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