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人気のある三角関数 >

2*sin(a)=sin(3a)

解

2 \cdot sin (a) = sin (3 a)

解

a = 2 πn, a = π + 2 πn, a = \frac{7 π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn, a = \frac{π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn

+1

度

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n, a = 21 0^{\circ} + 36 0^{\circ} n, a = 33 0^{\circ} + 36 0^{\circ} n, a = 3 0^{\circ} + 36 0^{\circ} n, a = 15 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin (a) = sin (3 a)

両辺から

sin (3 a)

を引く

2 sin (a) - sin (3 a) = 0

三角関数の公式を使用して書き換える

- sin (3 a) + 2 sin (a)

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

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)

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

簡素化

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

- (3 sin (a) - 4 sin^{3} (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))

括弧を分配する

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

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

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

類似した元を足す:

- 3 sin (a) + 2 sin (a) = - sin (a)

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

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

- sin (a) + 4 sin^{3} (a) = 0

置換で解く

- sin (a) + 4 sin^{3} (a) = 0

仮定:

sin (a) = u

- u + 4 u^{3} = 0

- u + 4 u^{3} = 0 : u = 0, u = - \frac{1}{2}, u = \frac{1}{2}

- u + 4 u^{3} = 0

因数

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

- u + 4 u^{3}

共通項をくくり出す

u : u (4 u^{2} - 1)

4 u^{3} - u

指数の規則を適用する:

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

u^{3} = u^{2} u

= 4 u^{2} u - u

共通項をくくり出す

u

= u (4 u^{2} - 1)

= u (4 u^{2} - 1)

因数

4 u^{2} - 1 : (2 u + 1) (2 u - 1)

4 u^{2} - 1

4 u^{2} - 1

を書き換え

(2 u)^{2} - 1^{2}

4 u^{2} - 1

4

を書き換え

2^{2}

= 2^{2} u^{2} - 1

1

を書き換え

1^{2}

= 2^{2} u^{2} - 1^{2}

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

2^{2} u^{2} = (2 u)^{2}

= (2 u)^{2} - 1^{2}

= (2 u)^{2} - 1^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

(2 u)^{2} - 1^{2} = (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1)

= u (2 u + 1) (2 u - 1)

u (2 u + 1) (2 u - 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

u = 0 or 2 u + 1 = 0 or 2 u - 1 = 0

解く

2 u + 1 = 0 : u = - \frac{1}{2}

2 u + 1 = 0

1

を右側に移動します

2 u + 1 = 0

両辺から

1

を引く

2 u + 1 - 1 = 0 - 1

簡素化

2 u = - 1

2 u = - 1

以下で両辺を割る

2

2 u = - 1

以下で両辺を割る

2

\frac{2 u}{2} = \frac{- 1}{2}

簡素化

u = - \frac{1}{2}

u = - \frac{1}{2}

解く

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

1

を右側に移動します

2 u - 1 = 0

両辺に

1

を足す

2 u - 1 + 1 = 0 + 1

簡素化

2 u = 1

2 u = 1

以下で両辺を割る

2

2 u = 1

以下で両辺を割る

2

\frac{2 u}{2} = \frac{1}{2}

簡素化

u = \frac{1}{2}

u = \frac{1}{2}

解答は

u = 0, u = - \frac{1}{2}, u = \frac{1}{2}

代用を戻す

u = sin (a)

sin (a) = 0, sin (a) = - \frac{1}{2}, sin (a) = \frac{1}{2}

sin (a) = 0, sin (a) = - \frac{1}{2}, sin (a) = \frac{1}{2}

sin (a) = 0 : a = 2 πn, a = π + 2 πn

sin (a) = 0

以下の一般解

sin (a) = 0

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = 0 + 2 πn, a = π + 2 πn

a = 0 + 2 πn, a = π + 2 πn

解く

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn, a = π + 2 πn

sin (a) = - \frac{1}{2} : a = \frac{7 π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

sin (a) = - \frac{1}{2}

以下の一般解

sin (a) = - \frac{1}{2}

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = \frac{7 π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

a = \frac{7 π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

sin (a) = \frac{1}{2} : a = \frac{π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn

sin (a) = \frac{1}{2}

以下の一般解

sin (a) = \frac{1}{2}

sin (x)

2 πn

循環を含む周期性テーブル:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

a = \frac{π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn

a = \frac{π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn

すべての解を組み合わせる

a = 2 πn, a = π + 2 πn, a = \frac{7 π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn, a = \frac{π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn

グラフ

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