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人気のある三角関数 >

sin^4(θ)=(cos^2(θ))/2

解

sin^{4} (θ) = \frac{cos ^{2} ( θ )}{2}

解

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

+1

度

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

sin^{4} (θ) = \frac{cos ^{2} ( θ )}{2}

両辺から

\frac{cos ^{2} ( θ )}{2}

を引く

sin^{4} (θ) - \frac{cos ^{2} ( θ )}{2} = 0

簡素化

sin^{4} (θ) - \frac{cos ^{2} ( θ )}{2} : \frac{2 sin ^{4} ( θ ) - cos ^{2} ( θ )}{2}

sin^{4} (θ) - \frac{cos ^{2} ( θ )}{2}

元を分数に変換する:

sin^{4} (θ) = \frac{sin ^{4} ( θ ) 2}{2}

= \frac{sin ^{4} ( θ ) \cdot 2}{2} - \frac{cos ^{2} ( θ )}{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 ^{4} ( θ ) \cdot 2 - cos ^{2} ( θ )}{2}

\frac{2 sin ^{4} ( θ ) - cos ^{2} ( θ )}{2} = 0

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

2 sin^{4} (θ) - cos^{2} (θ) = 0

因数

2 sin^{4} (θ) - cos^{2} (θ) : (2 sin^{2} (θ) + cos (θ)) (2 sin^{2} (θ) - cos (θ))

2 sin^{4} (θ) - cos^{2} (θ)

2 sin^{4} (θ) - cos^{2} (θ)

を書き換え

(2 sin^{2} (θ))^{2} - cos^{2} (θ)

2 sin^{4} (θ) - cos^{2} (θ)

累乗根の規則を適用する:

a = (a)^{2}

2 = (2)^{2}

= (2)^{2} sin^{4} (θ) - cos^{2} (θ)

指数の規則を適用する:

a^{b c} = (a^{b})^{c}

sin^{4} (θ) = (sin^{2} (θ))^{2}

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

指数の規則を適用する:

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

(2)^{2} (sin^{2} (θ))^{2} = (2 sin^{2} (θ))^{2}

= (2 sin^{2} (θ))^{2} - cos^{2} (θ)

= (2 sin^{2} (θ))^{2} - cos^{2} (θ)

2乗の差の公式を適用する:

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

(2 sin^{2} (θ))^{2} - cos^{2} (θ) = (2 sin^{2} (θ) + cos (θ)) (2 sin^{2} (θ) - cos (θ))

= (2 sin^{2} (θ) + cos (θ)) (2 sin^{2} (θ) - cos (θ))

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

各部分を別個に解く

2 sin^{2} (θ) + cos (θ) = 0 or 2 sin^{2} (θ) - cos (θ) = 0

2 sin^{2} (θ) + cos (θ) = 0 : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 sin^{2} (θ) + cos (θ) = 0

三角関数の公式を使用して書き換える

cos (θ) + sin^{2} (θ) 2

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)

= cos (θ) + (1 - cos^{2} (θ)) 2

cos (θ) + (1 - cos^{2} (θ)) 2 = 0

置換で解く

cos (θ) + (1 - cos^{2} (θ)) 2 = 0

仮定:

cos (θ) = u

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

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

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

拡張

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

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

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

拡張

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

2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

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

乗算:

1 \cdot 2 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

規則を適用

- (- a) = a

= 1 + 422

422 = 8

422

累乗根の規則を適用する:

a a = a

22 = 2

= 4 \cdot 2

数を乗じる:

4 \cdot 2 = 8

= 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{2}{2}

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

括弧を削除する:

(- a) = - a

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

数を足す/引く:

- 1 + 3 = 2

= \frac{2}{- 2 2}

分数の規則を適用する:

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

= - \frac{2}{2 2}

数を割る:

\frac{2}{2} = 1

= - \frac{1}{2}

有理化する

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

共役で乗じる

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

括弧を削除する:

(- a) = - a

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

数を引く:

- 1 - 3 = - 4

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

分数の規則を適用する:

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

= \frac{4}{2 2}

数を割る:

\frac{4}{2} = 2

= \frac{2}{2}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

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

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

指数の規則を適用する:

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

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

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

数を引く:

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

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

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

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

= 2

二次equationの解:

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

代用を戻す

u = cos (θ)

cos (θ) = - \frac{2}{2}, cos (θ) = 2

cos (θ) = - \frac{2}{2}, cos (θ) = 2

cos (θ) = - \frac{2}{2} : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = - \frac{2}{2}

以下の一般解

cos (θ) = - \frac{2}{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}

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = 2 :

解なし

cos (θ) = 2

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

2 sin^{2} (θ) - cos (θ) = 0 : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

2 sin^{2} (θ) - cos (θ) = 0

三角関数の公式を使用して書き換える

- cos (θ) + sin^{2} (θ) 2

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)

= - cos (θ) + (1 - cos^{2} (θ)) 2

- cos (θ) + (1 - cos^{2} (θ)) 2 = 0

置換で解く

- cos (θ) + (1 - cos^{2} (θ)) 2 = 0

仮定:

cos (θ) = u

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

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

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

拡張

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

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

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

拡張

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

2 (1 - u^{2})

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

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

乗算:

1 \cdot 2 = 2

= 2 - 2 u^{2}

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

= (- 1)^{2} + 422

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

422 = 8

422

累乗根の規則を適用する:

a a = a

22 = 2

= 4 \cdot 2

数を乗じる:

4 \cdot 2 = 8

= 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 )} : - 2

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

括弧を削除する:

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

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

数を足す:

1 + 3 = 4

= \frac{4}{- 2 2}

分数の規則を適用する:

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

= - \frac{4}{2 2}

数を割る:

\frac{4}{2} = 2

= \frac{2}{2}

累乗根の規則を適用する:

n a = a^{\frac{1}{n}}

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

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

指数の規則を適用する:

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

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

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

数を引く:

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

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

累乗根の規則を適用する:

a^{\frac{1}{n}} = n a

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

= - 2

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

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

括弧を削除する:

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

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

数を引く:

1 - 3 = - 2

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

分数の規則を適用する:

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

= \frac{2}{2 2}

数を割る:

\frac{2}{2} = 1

= \frac{1}{2}

有理化する

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

共役で乗じる

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

二次equationの解:

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

代用を戻す

u = cos (θ)

cos (θ) = - 2, cos (θ) = \frac{2}{2}

cos (θ) = - 2, cos (θ) = \frac{2}{2}

cos (θ) = - 2 :

解なし

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

解なし

cos (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

cos (θ) = \frac{2}{2}

以下の一般解

cos (θ) = \frac{2}{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}

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

グラフ

人気の例

2cos^2(x)+3cos(x)=-1

2 cos^{2} (x) + 3 cos (x) = - 1

2 cos (3 x) = 1

2sin(x)-sin(2x)=0

2 sin (x) - sin (2 x) = 0

2cos(x/2)-sqrt(2)=0

2 cos (\frac{x}{2}) - 2 = 0

cos (t) = 0