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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 cos(4B)=1-8sin^2(B)+8sin^4(B)

p ro v e cos (4 B) = 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

証明する sin(pi/2-u)=cos(u)

p ro v e sin (\frac{π}{2} - u) = cos (u)

証明する arccos(x)= 1/(cos(x))

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

証明する (csc^2(x)-1)/(cos(x))=cot(x)csc(x)

p ro v e \frac{csc ^{2} ( x ) - 1}{cos ( x )} = cot (x) csc (x)

証明する cos^2(x)(1-sec^2(x))=-sin^2(x)

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