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人気のある三角関数 >

証明する cos^3(x)= 3/4 cos(x)+1/4 cos(3x)

解

証明する

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

解

真

解答ステップ

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

右側を操作する

\frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

三角関数の公式を使用して書き換える

\frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

次の恒等を使用する:

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

cos (3 x)

三角関数の公式を使用して書き換える

cos (3 x)

書き換え

= cos (2 x + x)

角の和の公式を使用する:

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

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

2倍角の公式を使用:

sin (2 x) = 2 sin (x) cos (x)

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

簡素化

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

2 sin (x) cos (x) sin (x)

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

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

2倍角の公式を使用:

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

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

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)

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

拡張

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

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

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

2 cos^{2} (x) cos (x)

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

乗算:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

マイナス・プラスの規則を適用する

- (- a) = a

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

簡素化

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

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

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

2 \cdot 1 cos (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos (x)

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

2 cos^{2} (x) cos (x)

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

類似した元を足す:

- cos (x) - 2 cos (x) = - 3 cos (x)

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

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

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

= \frac{3}{4} cos (x) + \frac{1}{4} (4 cos^{3} (x) - 3 cos (x))

簡素化

\frac{3}{4} cos (x) + \frac{1}{4} (4 cos^{3} (x) - 3 cos (x)) : cos^{3} (x)

\frac{3}{4} cos (x) + \frac{1}{4} (4 cos^{3} (x) - 3 cos (x))

拡張

\frac{1}{4} (4 cos^{3} (x) - 3 cos (x)) : cos^{3} (x) - \frac{3}{4} cos (x)

\frac{1}{4} (4 cos^{3} (x) - 3 cos (x))

分配法則を適用する:

a (b - c) = ab - a c

a = \frac{1}{4}, b = 4 cos^{3} (x), c = 3 cos (x)

= \frac{1}{4} \cdot 4 cos^{3} (x) - \frac{1}{4} \cdot 3 cos (x)

= 4 \cdot \frac{1}{4} cos^{3} (x) - 3 \cdot \frac{1}{4} cos (x)

簡素化

4 \cdot \frac{1}{4} cos^{3} (x) - 3 \cdot \frac{1}{4} cos (x) : cos^{3} (x) - \frac{3}{4} cos (x)

4 \cdot \frac{1}{4} cos^{3} (x) - 3 \cdot \frac{1}{4} cos (x)

4 \cdot \frac{1}{4} cos^{3} (x) = cos^{3} (x)

4 \cdot \frac{1}{4} cos^{3} (x)

分数を乗じる:

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

= \frac{1 \cdot 4}{4} cos^{3} (x)

共通因数を約分する:

4

= cos^{3} (x) \cdot 1

乗算:

cos^{3} (x) \cdot 1 = cos^{3} (x)

= cos^{3} (x)

3 \cdot \frac{1}{4} cos (x) = \frac{3}{4} cos (x)

3 \cdot \frac{1}{4} cos (x)

分数を乗じる:

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

= \frac{1 \cdot 3}{4} cos (x)

数を乗じる:

1 \cdot 3 = 3

= \frac{3}{4} cos (x)

= cos^{3} (x) - \frac{3}{4} cos (x)

= cos^{3} (x) - \frac{3}{4} cos (x)

= \frac{3}{4} cos (x) + cos^{3} (x) - \frac{3}{4} cos (x)

簡素化

\frac{3}{4} cos (x) + cos^{3} (x) - \frac{3}{4} cos (x) : cos^{3} (x)

\frac{3}{4} cos (x) + cos^{3} (x) - \frac{3}{4} cos (x)

条件のようなグループ

= \frac{3}{4} cos (x) - \frac{3}{4} cos (x) + cos^{3} (x)

類似した元を足す:

\frac{3}{4} cos (x) - \frac{3}{4} cos (x) = 0

\frac{3}{4} cos (x) - \frac{3}{4} cos (x)

共通項をくくり出す

cos (x)

= cos (x) (\frac{3}{4} - \frac{3}{4})

\frac{3}{4} - \frac{3}{4} = 0

\frac{3}{4} - \frac{3}{4}

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{3 - 3}{4}

改良

= 0

= 0

= cos^{3} (x)

= cos^{3} (x)

= cos^{3} (x)

= cos^{3} (x)

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 tan(pi/4+x)=(1+sin(2x))/(cos(2x))

p ro v e tan (\frac{π}{4} + x) = \frac{1 + sin ( 2 x )}{cos ( 2 x )}

証明する sin^2(x)+1/(sec^2(x))=sin(x)csc(x)

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

証明する sin(x)=sqrt(1-cos(x))

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

証明する (sec(2x)+1)/(tan(2x))=cot(x)

p ro v e \frac{sec ( 2 x ) + 1}{tan ( 2 x )} = cot (x)

証明する cos^2(6θ)-sin^2(6θ)=cos(12θ)

p ro v e cos^{2} (6 θ) - sin^{2} (6 θ) = cos (12 θ)