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人気のある三角関数 >

1-sin^2(x)-cos(2x)= 1/2

解

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

解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = - 13 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

\frac{1}{2}

を引く

- sin^{2} (x) - cos (2 x) + \frac{1}{2} = 0

簡素化

- sin^{2} (x) - cos (2 x) + \frac{1}{2} : \frac{- 2 sin ^{2} ( x ) - 2 cos ( 2 x ) + 1}{2}

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

元を分数に変換する:

sin^{2} (x) = \frac{sin ^{2} ( x ) 2}{2}, cos (2 x) = \frac{cos ( 2 x ) 2}{2}

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

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

\frac{- 2 sin ^{2} ( x ) - 2 cos ( 2 x ) + 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 2 sin^{2} (x) - 2 cos (2 x) + 1 = 0

三角関数の公式を使用して書き換える

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

2倍角の公式を使用:

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

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

類似した元を足す:

- 2 cos (2 x) + 2 cos (2 x) = 0

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

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

1 - 2 cos^{2} (x) = 0

置換で解く

1 - 2 cos^{2} (x) = 0

仮定:

cos (x) = u

1 - 2 u^{2} = 0

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

1 - 2 u^{2} = 0

1

を右側に移動します

1 - 2 u^{2} = 0

両辺から

1

を引く

1 - 2 u^{2} - 1 = 0 - 1

簡素化

- 2 u^{2} = - 1

- 2 u^{2} = - 1

以下で両辺を割る

- 2

- 2 u^{2} = - 1

以下で両辺を割る

- 2

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

簡素化

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

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

x^{2} = f (a)

の場合, 解は

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

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

代用を戻す

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{2} : x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = \frac{1}{2}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{2}

以下の一般解

cos (x) = \frac{1}{2}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2} : x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

cos (x) = - \frac{1}{2}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1}{2}

以下の一般解

cos (x) = - \frac{1}{2}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn, x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

10進法形式で解を証明する

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

グラフ

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