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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 (1-sin(x))/(1-csc(x))=-sin(x)

p ro v e \frac{1 - sin ( x )}{1 - csc ( x )} = - sin (x)

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p ro v e 1 + cos (10 y) = 2 cos^{2} (5 y)

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p ro v e sin (\frac{π}{6}) = \frac{1}{2}

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

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