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人気のある三角関数 >

証明する cos(4a)=8cos^4(a)-8cos^2(a)+1

解

証明する

cos (4 a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

解

真

解答ステップ

cos (4 a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

左側を操作する

cos (4 a)

三角関数の公式を使用して書き換える

cos (4 a)

= cos (2 \cdot 2 a)

2倍角の公式を使用:

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

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

2倍角の公式を使用:

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

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

簡素化

2 (2 cos^{2} (a) - 1)^{2} - 1 : 8 cos^{4} (a) - 8 cos^{2} (a) + 1

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

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

完全平方式を適用する:

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

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

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

簡素化

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

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

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

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

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

指数の規則を適用する:

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

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

数を乗じる:

2 \cdot 2 = 4

= cos^{4} (a)

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

2^{2} = 4

= 4 cos^{4} (a)

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

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

数を乗じる:

2 \cdot 2 \cdot 1 = 4

= 4 cos^{2} (a)

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

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

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

拡張

2 (4 cos^{4} (a) - 4 cos^{2} (a) + 1) : 8 cos^{4} (a) - 8 cos^{2} (a) + 2

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

括弧を分配する

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

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

2 \cdot 4 cos^{4} (a) - 2 \cdot 4 cos^{2} (a) + 2 \cdot 1 : 8 cos^{4} (a) - 8 cos^{2} (a) + 2

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

数を乗じる:

2 \cdot 4 = 8

= 8 cos^{4} (a) - 8 cos^{2} (a) + 2 \cdot 1

数を乗じる:

2 \cdot 1 = 2

= 8 cos^{4} (a) - 8 cos^{2} (a) + 2

= 8 cos^{4} (a) - 8 cos^{2} (a) + 2

= 8 cos^{4} (a) - 8 cos^{2} (a) + 2 - 1

数を足す/引く:

2 - 1 = 1

= 8 cos^{4} (a) - 8 cos^{2} (a) + 1

= 8 cos^{4} (a) - 8 cos^{2} (a) + 1

= 8 cos^{4} (a) - 8 cos^{2} (a) + 1

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 2cos^2(x)+cos(x)=0

p ro v e 2 cos^{2} (x) + cos (x) = 0

証明する cos(2θ)=(1-tan^2(θ))/(sec^2(θ))

p ro v e cos (2 θ) = \frac{1 - tan ^{2} ( θ )}{sec ^{2} ( θ )}

証明する sin^5(x)=(1-2cos^2(x)+cos^4(x))(sin(x))

p ro v e sin^{5} (x) = (1 - 2 cos^{2} (x) + cos^{4} (x)) (sin (x))

証明する sin^2(t/2)=(1-cos(t))/2

p ro v e sin^{2} (\frac{t}{2}) = \frac{1 - cos ( t )}{2}

証明する sec(2x)= 1/(cos(2x))

p ro v e sec (2 x) = \frac{1}{cos ( 2 x )}