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人気のある三角関数 >

証明する sin^6(a)+cos^6(a)=1-3sin^2(a)*cos^2(a)

解

証明する

sin^{6} (a) + cos^{6} (a) = 1 - 3 sin^{2} (a) \cdot cos^{2} (a)

解

真

解答ステップ

sin^{6} (a) + cos^{6} (a) = 1 - 3 sin^{2} (a) cos^{2} (a)

左側を操作する

sin^{6} (a) + cos^{6} (a)

因数

cos^{6} (a) + sin^{6} (a) : (cos^{2} (a) + sin^{2} (a)) (cos^{4} (a) - sin^{2} (a) cos^{2} (a) + sin^{4} (a))

cos^{6} (a) + sin^{6} (a)

cos^{6} (a) + sin^{6} (a)

を書き換え

(cos^{2} (a))^{3} + (sin^{2} (a))^{3}

cos^{6} (a) + sin^{6} (a)

指数の規則を適用する:

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

sin^{6} (a) = (sin^{2} (a))^{3}

= cos^{6} (a) + (sin^{2} (a))^{3}

指数の規則を適用する:

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

cos^{6} (a) = (cos^{2} (a))^{3}

= (cos^{2} (a))^{3} + (sin^{2} (a))^{3}

= (cos^{2} (a))^{3} + (sin^{2} (a))^{3}

立方数の和の公式を適用する:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(cos^{2} (a))^{3} + (sin^{2} (a))^{3} = (cos^{2} (a) + sin^{2} (a)) (cos^{4} (a) - sin^{2} (a) cos^{2} (a) + sin^{4} (a))

= (sin^{2} (a) + cos^{2} (a)) (sin^{4} (a) - sin^{2} (a) cos^{2} (a) + cos^{4} (a))

= (cos^{2} (a) + sin^{2} (a)) (cos^{4} (a) + sin^{4} (a) - cos^{2} (a) sin^{2} (a))

三角関数の公式を使用して書き換える

(cos^{2} (a) + sin^{2} (a)) (cos^{4} (a) + sin^{4} (a) - cos^{2} (a) sin^{2} (a))

ピタゴラスの公式を使用する:

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

= 1 \cdot (cos^{4} (a) + sin^{4} (a) - cos^{2} (a) sin^{2} (a))

簡素化

= cos^{4} (a) + sin^{4} (a) - cos^{2} (a) sin^{2} (a)

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)

= cos^{4} (a) + sin^{4} (a) + cos^{2} (a) (cos (2 a) - cos^{2} (a))

簡素化

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

cos^{4} (a) + sin^{4} (a) + cos^{2} (a) (cos (2 a) - cos^{2} (a))

拡張

cos^{2} (a) (cos (2 a) - cos^{2} (a)) : cos^{2} (a) cos (2 a) - cos^{4} (a)

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

cos^{2} (a) cos^{2} (a) = cos^{4} (a)

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

指数の規則を適用する:

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

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

= cos^{2 + 2} (a)

数を足す:

2 + 2 = 4

= cos^{4} (a)

= cos^{2} (a) cos (2 a) - cos^{4} (a)

= cos^{4} (a) + sin^{4} (a) + cos^{2} (a) cos (2 a) - cos^{4} (a)

類似した元を足す:

cos^{4} (a) - cos^{4} (a) = 0

= sin^{4} (a) + cos^{2} (a) cos (2 a)

= sin^{4} (a) + cos^{2} (a) cos (2 a)

2倍角の公式を使用:

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

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

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)

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

簡素化

sin^{4} (a) + (1 - sin^{2} (a)) (1 - 2 sin^{2} (a)) : 3 sin^{4} (a) - 3 sin^{2} (a) + 1

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

拡張

(1 - sin^{2} (a)) (1 - 2 sin^{2} (a)) : 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

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

FOIL メソッドを適用する:

(a + b) (c + d) = a c + a d + b c + b d

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

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

マイナス・プラスの規則を適用する

+ (- a) = - a, (- a) (- b) = ab

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

簡素化

1 \cdot 1 - 1 \cdot 2 sin^{2} (a) - 1 \cdot sin^{2} (a) + 2 sin^{2} (a) sin^{2} (a) : 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

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

1 \cdot 1 = 1

1 \cdot 1

数を乗じる:

1 \cdot 1 = 1

= 1

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

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

数を乗じる:

1 \cdot 2 = 2

= 2 sin^{2} (a)

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

1 \cdot sin^{2} (a)

乗算:

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

= sin^{2} (a)

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

2 sin^{2} (a) sin^{2} (a)

指数の規則を適用する:

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

sin^{2} (a) sin^{2} (a) = sin^{2 + 2} (a)

= 2 sin^{2 + 2} (a)

数を足す:

2 + 2 = 4

= 2 sin^{4} (a)

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

類似した元を足す:

- 2 sin^{2} (a) - sin^{2} (a) = - 3 sin^{2} (a)

= 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

= 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

= sin^{4} (a) + 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

簡素化

sin^{4} (a) + 1 - 3 sin^{2} (a) + 2 sin^{4} (a) : 3 sin^{4} (a) - 3 sin^{2} (a) + 1

sin^{4} (a) + 1 - 3 sin^{2} (a) + 2 sin^{4} (a)

条件のようなグループ

= sin^{4} (a) - 3 sin^{2} (a) + 2 sin^{4} (a) + 1

類似した元を足す:

sin^{4} (a) + 2 sin^{4} (a) = 3 sin^{4} (a)

= 3 sin^{4} (a) - 3 sin^{2} (a) + 1

= 3 sin^{4} (a) - 3 sin^{2} (a) + 1

= 3 sin^{4} (a) - 3 sin^{2} (a) + 1

= 3 sin^{4} (a) - 3 sin^{2} (a) + 1

因数

- 3 sin^{2} (a) + 3 sin^{4} (a) : 3 sin^{2} (a) (sin (a) + 1) (sin (a) - 1)

- 3 sin^{2} (a) + 3 sin^{4} (a)

共通項をくくり出す

3 sin^{2} (a) : 3 sin^{2} (a) (sin^{2} (a) - 1)

3 sin^{4} (a) - 3 sin^{2} (a)

指数の規則を適用する:

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

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

= 3 sin^{2} (a) sin^{2} (a) - 3 sin^{2} (a)

共通項をくくり出す

3 sin^{2} (a)

= 3 sin^{2} (a) (sin^{2} (a) - 1)

= 3 sin^{2} (a) (sin^{2} (a) - 1)

因数

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

sin^{2} (a) - 1

1

を書き換え

1^{2}

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

2乗の差の公式を適用する:

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

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

= (sin (a) + 1) (sin (a) - 1)

= 3 sin^{2} (a) (sin (a) + 1) (sin (a) - 1)

= 1 + (- 1 + sin (a)) (1 + sin (a)) \cdot 3 sin^{2} (a)

三角関数の公式を使用して書き換える

1 + (- 1 + sin (a)) (1 + sin (a)) \cdot 3 sin^{2} (a)

拡張

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

(sin (a) + 1) (sin (a) - 1)

2乗の差の公式を適用する:

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

a = sin (a), b = 1

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

規則を適用

1^{a} = 1

1^{2} = 1

= sin^{2} (a) - 1

= 1 + 3 sin^{2} (a) (sin^{2} (a) - 1)

ピタゴラスの公式を使用する:

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

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

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

簡素化

= 1 - 3 sin^{2} (a) cos^{2} (a)

= 1 - 3 sin^{2} (a) cos^{2} (a)

両辺を同じ形式にできることを証明した

\Rightarrow 真

人気の例

証明する sin(-3x)=-sin(3x)

p ro v e sin (- 3 x) = - sin (3 x)

証明する cos(x)+sin(x)=tan(x)+1

p ro v e cos (x) + sin (x) = tan (x) + 1

証明する csc(2θ)=((csc(θ)sec(θ)))/2

p ro v e csc (2 θ) = \frac{( csc ( θ ) sec ( θ ) )}{2}

証明する (sin(a)+cos(a))/(cos(a))=tan(a)+1

p ro v e \frac{sin ( a ) + cos ( a )}{cos ( a )} = tan (a) + 1

証明する cos(5*pi)=-1

p ro v e cos (5 \cdot π) = - 1