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人気のある三角関数 >

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

解

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

解

a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn, a = 2 πn

+1

度

a = 12 0^{\circ} + 36 0^{\circ} n, a = 24 0^{\circ} + 36 0^{\circ} n, a = 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

\frac{1 - cos ( a )}{2}

を引く

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

簡素化

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

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

元を分数に変換する:

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

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

拡張

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

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

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

- (1 - cos (a)) : - 1 + cos (a)

- (1 - cos (a))

括弧を分配する

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

マイナス・プラスの規則を適用する

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

= - 1 + cos (a)

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

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

拡張

2 (1 - cos^{2} (a)) : 2 - 2 cos^{2} (a)

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

分配法則を適用する:

a (b - c) = ab - a c

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

= 2 \cdot 1 - 2 cos^{2} (a)

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

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

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

条件のようなグループ

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

数を足す/引く:

- 1 + 2 = 1

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

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

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

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

置換で解く

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

仮定:

cos (a) = u

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

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

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

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 1 = 3

1^{2} - 4 (- 2) \cdot 1

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

数を足す/引く:

- 1 + 3 = 2

= \frac{2}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{- 4}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

数を引く:

- 1 - 3 = - 4

= \frac{- 4}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

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

代用を戻す

u = cos (a)

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

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

cos (a) = - \frac{1}{2} : a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

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

以下の一般解

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

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 = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn

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

すべての解を組み合わせる

a = \frac{2 π}{3} + 2 πn, a = \frac{4 π}{3} + 2 πn, a = 2 πn

グラフ

人気の例

cos(x)=(-11)/(12)

cos (x) = \frac{- 11}{12}

6sin^2(x)+5cos(x)-2=0

6 sin^{2} (x) + 5 cos (x) - 2 = 0

(sin^{22}(a))/(sin^2(a))=4-4sin^2(a)

\frac{sin ^{22} ( a )}{sin ^{2} ( a )} = 4 - 4 sin^{2} (a)

tan^{3} (x) = 2

sin^3(x)=3sin(x)

sin^{3} (x) = 3 sin (x)