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人気のある三角関数 >

cos^4(x)-sin^4(x)=1

解

cos^{4} (x) - sin^{4} (x) = 1

解

x = 2 πn, x = π + 2 πn

+1

度

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

解答ステップ

cos^{4} (x) - sin^{4} (x) = 1

両辺から

1

を引く

cos^{4} (x) - sin^{4} (x) - 1 = 0

指数の規則を適用する:

a^{b} = a^{2} a^{b - 2}

- 1 - sin^{4} (x) + cos^{2} (x) cos^{2} (x) = 0

三角関数の公式を使用して書き換える

- 1 - sin^{4} (x) + cos^{2} (x) cos^{2} (x)

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 - sin^{4} (x) + (1 - sin^{2} (x)) (1 - sin^{2} (x))

簡素化

- 1 - sin^{4} (x) + (1 - sin^{2} (x)) (1 - sin^{2} (x)) : - 2 sin^{2} (x)

- 1 - sin^{4} (x) + (1 - sin^{2} (x)) (1 - sin^{2} (x))

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

= - 1 - sin^{4} (x) + (- sin^{2} (x) + 1)^{2}

(1 - sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

完全平方式を適用する:

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

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

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

簡素化

1^{2} - 2 \cdot 1 \cdot sin^{2} (x) + (sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

数を乗じる:

2 \cdot 1 = 2

= 2 sin^{2} (x)

(sin^{2} (x))^{2} = sin^{4} (x)

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

指数の規則を適用する:

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

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

数を乗じる:

2 \cdot 2 = 4

= sin^{4} (x)

= 1 - 2 sin^{2} (x) + sin^{4} (x)

= 1 - 2 sin^{2} (x) + sin^{4} (x)

= - 1 - sin^{4} (x) + 1 - 2 sin^{2} (x) + sin^{4} (x)

簡素化

- 1 - sin^{4} (x) + 1 - 2 sin^{2} (x) + sin^{4} (x) : - 2 sin^{2} (x)

- 1 - sin^{4} (x) + 1 - 2 sin^{2} (x) + sin^{4} (x)

条件のようなグループ

= - sin^{4} (x) - 2 sin^{2} (x) + sin^{4} (x) - 1 + 1

類似した元を足す:

- sin^{4} (x) + sin^{4} (x) = 0

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

- 1 + 1 = 0

= - 2 sin^{2} (x)

= - 2 sin^{2} (x)

= - 2 sin^{2} (x)

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

以下で両辺を割る

- 2

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

以下で両辺を割る

- 2

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

以下で両辺を割る

- 2

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

簡素化

sin^{2} (x) = 0

sin^{2} (x) = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

以下の一般解

sin (x) = 0

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

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

グラフ

人気の例

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4 sec^{2} (x) tan (x) = 8 tan (x)

sin(x)(sin(x)-1)=0

sin (x) (sin (x) - 1) = 0

tan (2 x) = tan (x)

tan^{2} (x) = - 1

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