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人気のある三角関数 >

証明する sin^4(u)-cos^4(u)=2sin^2(u)-1

解

証明する

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

解

真

解答ステップ

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

左側を操作する

sin^{4} (u) - cos^{4} (u)

因数

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

- cos^{4} (u) + sin^{4} (u)

sin^{4} (u) - cos^{4} (u)

を書き換え

(sin^{2} (u))^{2} - (cos^{2} (u))^{2}

sin^{4} (u) - cos^{4} (u)

指数の規則を適用する:

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

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

= (sin^{2} (u))^{2} - cos^{4} (u)

指数の規則を適用する:

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

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

= (sin^{2} (u))^{2} - (cos^{2} (u))^{2}

= (sin^{2} (u))^{2} - (cos^{2} (u))^{2}

2乗の差の公式を適用する:

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

(sin^{2} (u))^{2} - (cos^{2} (u))^{2} = (sin^{2} (u) + cos^{2} (u)) (sin^{2} (u) - cos^{2} (u))

= (sin^{2} (u) + cos^{2} (u)) (sin^{2} (u) - cos^{2} (u))

因数

sin^{2} (u) - cos^{2} (u) : (sin (u) + cos (u)) (sin (u) - cos (u))

sin^{2} (u) - cos^{2} (u)

2乗の差の公式を適用する:

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

sin^{2} (u) - cos^{2} (u) = (sin (u) + cos (u)) (sin (u) - cos (u))

= (sin (u) + cos (u)) (sin (u) - cos (u))

= (sin^{2} (u) + cos^{2} (u)) (sin (u) + cos (u)) (sin (u) - cos (u))

= (- cos (u) + sin (u)) (cos (u) + sin (u)) (cos^{2} (u) + sin^{2} (u))

三角関数の公式を使用して書き換える

(- cos (u) + sin (u)) (cos (u) + sin (u)) (cos^{2} (u) + sin^{2} (u))

拡張

(sin (u) + cos (u)) (sin (u) - cos (u)) : sin^{2} (u) - cos^{2} (u)

(sin (u) + cos (u)) (sin (u) - cos (u))

2乗の差の公式を適用する:

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

a = sin (u), b = cos (u)

= sin^{2} (u) - cos^{2} (u)

= (sin^{2} (u) - cos^{2} (u)) (cos^{2} (u) + sin^{2} (u))

2倍角の公式を使用:

cos^{2} (u) - sin^{2} (u) = cos (2 u)

- cos^{2} (u) + sin^{2} (u) = - cos (2 u)

= (- cos (2 u)) (cos^{2} (u) + sin^{2} (u))

簡素化

= - cos (2 u) (cos^{2} (u) + sin^{2} (u))

2倍角の公式を使用:

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

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

ピタゴラスの公式を使用する:

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

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

簡素化

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

- (1 - 2 sin^{2} (u)) \cdot 1

乗算:

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

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

括弧を分配する

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

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

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

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

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

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

証明する (tan(x)+cot(x))tan(x)=sec^2(x)

p ro v e (tan (x) + cot (x)) tan (x) = sec^{2} (x)

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

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

証明する 1-cos^2(x)=(tan^2(x))/(sec^2(x))

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

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

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