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人気のある三角関数 >

証明する sin(3A)=3sin(A)-4sin^3(A)

解

証明する

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

解

真

解答ステップ

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

左側を操作する

sin (3 A)

三角関数の公式を使用して書き換える

sin (3 A)

書き換え

= sin (2 A + A)

角の和の公式を使用する:

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

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

2倍角の公式を使用:

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

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

簡素化

cos (2 A) sin (A) + cos (A) \cdot 2 sin (A) cos (A) : sin (A) cos (2 A) + 2 cos^{2} (A) sin (A)

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

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

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

2倍角の公式を使用:

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

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

ピタゴラスの公式を使用する:

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

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

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

拡張

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

(1 - 2 sin^{2} (A)) sin (A) + 2 (1 - sin^{2} (A)) sin (A)

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

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

1 \cdot sin (A) = sin (A)

1 sin (A)

乗算:

1 \cdot sin (A) = sin (A)

= sin (A)

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 sin^{3} (A)

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

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

= sin (A) - 2 sin^{3} (A) + 2 (1 - sin^{2} (A)) sin (A)

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

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

2 \cdot 1 sin (A) - 2 sin^{2} (A) sin (A)

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

2 \cdot 1 sin (A)

数を乗じる:

2 \cdot 1 = 2

= 2 sin (A)

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 sin^{3} (A)

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

類似した元を足す:

sin (A) + 2 sin (A) = 3 sin (A)

= - 4 sin^{3} (A) + 3 sin (A)

= - 4 sin^{3} (A) + 3 sin (A)

= - 4 sin^{3} (A) + 3 sin (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 cot^4(x)+cot^2(x)=cot^2(x)csc^2(x)

p ro v e cot^{4} (x) + cot^{2} (x) = cot^{2} (x) csc^{2} (x)

証明する cot^3(x)=cot(x)(csc^2(x)-1)

p ro v e cot^{3} (x) = cot (x) (csc^{2} (x) - 1)

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

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

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