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人気のある三角関数 >

証明する cos^4(A)-sin^4(A)+1=2cos^2(A)

解

証明する

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

解

真

解答ステップ

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

左側を操作する

cos^{4} (A) - sin^{4} (A) + 1

因数

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

1 - sin^{4} (A)

共通項をくくり出す

- 1

= - (sin^{4} (A) - 1)

因数

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

sin^{4} (A) - 1

sin^{4} (A) - 1

を書き換え

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

sin^{4} (A) - 1

1

を書き換え

1^{2}

= sin^{4} (A) - 1^{2}

指数の規則を適用する:

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

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

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

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

= (sin^{2} (A) + 1) (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)

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

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

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

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)

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

簡素化

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

cos^{4} (A) - (- cos^{2} (A)) (2 - cos^{2} (A))

規則を適用

- (- a) = a

= cos^{4} (A) + cos^{2} (A) (2 - cos^{2} (A))

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

= cos^{2} (A) \cdot 2 - cos^{2} (A) cos^{2} (A)

= 2 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)

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

= cos^{4} (A) + 2 cos^{2} (A) - cos^{4} (A)

類似した元を足す:

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

= 2 cos^{2} (A)

= 2 cos^{2} (A)

= 2 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

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p ro v e 1 - tan^{4} (θ) = 2 sec^{2} (θ) - sec^{4} (θ)

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p ro v e 7 sec (y) cos (y) = 7

証明する sec(pi/2-u)=csc(u)

p ro v e sec (\frac{π}{2} - u) = csc (u)

証明する (cot^3(t))/(csc(t))=cos(t)(csc^2(-1))

p ro v e \frac{cot ^{3} ( t )}{csc ( t )} = cos (t) (csc^{2} (- 1))

証明する (sec^2(x)cot(x))/(csc^2(x))=tan(x)

p ro v e \frac{sec ^{2} ( x ) cot ( x )}{csc ^{2} ( x )} = tan (x)