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人気のある三角関数 >

(2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1

解

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

解

a = 2 πn, a = π + 2 πn

+1

度

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

置換で解く

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

仮定:

cos (a) = u

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

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

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

拡張

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

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

2乗の差の公式を適用する:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

簡素化

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

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

(2 u)^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} u^{2}

2^{2} = 4

= 4 u^{2}

= 4 u^{2} - 1

= 4 u^{2} - 1

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

1

を右側に移動します

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

両辺に

1

を足す

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

簡素化

4 u^{2} = 2 u^{2} + 2

4 u^{2} = 2 u^{2} + 2

2 u^{2}

を左側に移動します

4 u^{2} = 2 u^{2} + 2

両辺から

2 u^{2}

を引く

4 u^{2} - 2 u^{2} = 2 u^{2} + 2 - 2 u^{2}

簡素化

2 u^{2} = 2

2 u^{2} = 2

以下で両辺を割る

2

2 u^{2} = 2

以下で両辺を割る

2

\frac{2 u ^{2}}{2} = \frac{2}{2}

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = cos (a)

cos (a) = 1, cos (a) = - 1

cos (a) = 1, cos (a) = - 1

cos (a) = 1 : a = 2 πn

cos (a) = 1

以下の一般解

cos (a) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 0 + 2 πn

a = 0 + 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

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

以下の一般解

cos (a) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = π + 2 πn

a = π + 2 πn

すべての解を組み合わせる

a = 2 πn, a = π + 2 πn

グラフ

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