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人気のある >

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

解

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

解

x = 1.17809 \dots + 2 πn, x = π - 1.17809 \dots + 2 πn, x = - 1.17809 \dots + 2 πn, x = π + 1.17809 \dots + 2 πn

+1

度

x = 67. 5^{\circ} + 36 0^{\circ} n, x = 112. 5^{\circ} + 36 0^{\circ} n, x = - 67. 5^{\circ} + 36 0^{\circ} n, x = 247. 5^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

sin^{2} (x) - \frac{2}{2}

を引く

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

簡素化

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

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

元を分数に変換する:

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

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

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

1 + (1 - sin^{2} (x)) 2 - sin^{2} (x) 2 : 1 + 2 - 22 sin^{2} (x)

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

乗算:

1 \cdot 2 = 2

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

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

類似した元を足す:

- 2 sin^{2} (x) - 2 sin^{2} (x) = - 22 sin^{2} (x)

= 1 + 2 - 22 sin^{2} (x)

= 1 + 2 - 22 sin^{2} (x)

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

置換で解く

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

仮定:

sin (x) = u

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

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

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

1

を右側に移動します

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

両辺から

1

を引く

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

簡素化

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

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

2

を右側に移動します

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

両辺から

2

を引く

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

簡素化

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

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

以下で両辺を割る

- 22

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

以下で両辺を割る

- 22

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

簡素化

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

簡素化

\frac{- 2 u ^{2} 2}{- 2 2} : u^{2}

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

分数の規則を適用する:

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

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

数を割る:

\frac{2}{2} = 1

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

共通因数を約分する:

2

= u^{2}

簡素化

- \frac{1}{- 2 2} - \frac{2}{- 2 2} : \frac{2 + 2}{4}

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

規則を適用

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

分数の規則を適用する:

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

- 1 - 2 = - (1 + 2)

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

有理化する

\frac{1 + 2}{2 2} : \frac{2 + 2}{2 ^{2}}

\frac{1 + 2}{2 2}

共役で乗じる

\frac{2}{2}

= \frac{( 1 + 2 ) 2}{2 2 2}

(1 + 2) 2 = 2 + 2

(1 + 2) 2

= 2 (1 + 2)

分配法則を適用する:

a (b + c) = ab + a c

a = 2, b = 1, c = 2

= 2 \cdot 1 + 22

= 1 \cdot 2 + 22

簡素化

1 \cdot 2 + 22 : 2 + 2

1 \cdot 2 + 22

乗算:

1 \cdot 2 = 2

= 2 + 22

累乗根の規則を適用する:

a a = a

22 = 2

= 2 + 2

= 2 + 2

222 = 2^{2}

222

指数の規則を適用する:

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

222 = 2 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 2^{1 + \frac{1}{2} + \frac{1}{2}}

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

類似した元を足す:

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

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

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

2 \cdot \frac{1}{2}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

共通因数を約分する:

2

= 1

= 2^{1 + 1}

数を足す:

1 + 1 = 2

= 2^{2}

= \frac{2 + 2}{2 ^{2}}

= \frac{2 + 2}{2 ^{2}}

2^{2} = 4

= \frac{2 + 2}{4}

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

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

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

x^{2} = f (a)

の場合, 解は

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

u = \frac{2 + 2}{4}, u = - \frac{2 + 2}{4}

\frac{2 + 2}{4} = \frac{2 + 2}{2}

\frac{2 + 2}{4}

累乗根の規則を適用する:

, 以下を想定

a \geq 0, b \geq 0

= \frac{2 + 2}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

2^{2} = 2

= 2

= \frac{2 + 2}{2}

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

- \frac{2 + 2}{4}

簡素化

\frac{2 + 2}{4} : \frac{2 + 2}{2}

\frac{2 + 2}{4}

累乗根の規則を適用する:

, 以下を想定

a \geq 0, b \geq 0

= \frac{2 + 2}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

2^{2} = 2

= 2

= \frac{2 + 2}{2}

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

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

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

代用を戻す

u = sin (x)

sin (x) = \frac{2 + 2}{2}, sin (x) = - \frac{2 + 2}{2}

sin (x) = \frac{2 + 2}{2}, sin (x) = - \frac{2 + 2}{2}

sin (x) = \frac{2 + 2}{2} : x = arcsin (\frac{2 + 2}{2}) + 2 πn, x = π - arcsin (\frac{2 + 2}{2}) + 2 πn

sin (x) = \frac{2 + 2}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{2 + 2}{2}

以下の一般解

sin (x) = \frac{2 + 2}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

sin (x) = - \frac{2 + 2}{2} : x = arcsin (- \frac{2 + 2}{2}) + 2 πn, x = π + arcsin (\frac{2 + 2}{2}) + 2 πn

sin (x) = - \frac{2 + 2}{2}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{2 + 2}{2}

以下の一般解

sin (x) = - \frac{2 + 2}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.17809 \dots + 2 πn, x = π - 1.17809 \dots + 2 πn, x = - 1.17809 \dots + 2 πn, x = π + 1.17809 \dots + 2 πn

人気の例

cos(2θ)-sin(θ)=0

cos (2 θ) - sin (θ) = 0

tan(θ)=-(2sqrt(3))/3 sin(θ)

tan (θ) = - \frac{2 3}{3} sin (θ)

sqrt(3)cos(x)tan(x)-cos(x)=0

3 cos (x) tan (x) - cos (x) = 0

4 sin^{2} (θ) = 3

cos(2x)=sin(x)+1

cos (2 x) = sin (x) + 1