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人気のある三角関数 >

証明する sec^4(t)-tan^4(t)=1+2tan^2(t)

解

証明する

sec^{4} (t) - tan^{4} (t) = 1 + 2 tan^{2} (t)

解

真

解答ステップ

sec^{4} (t) - tan^{4} (t) = 1 + 2 tan^{2} (t)

左側を操作する

sec^{4} (t) - tan^{4} (t)

因数

sec^{4} (t) - tan^{4} (t) : (sec^{2} (t) + tan^{2} (t)) (sec (t) + tan (t)) (sec (t) - tan (t))

sec^{4} (t) - tan^{4} (t)

sec^{4} (t) - tan^{4} (t)

を書き換え

(sec^{2} (t))^{2} - (tan^{2} (t))^{2}

sec^{4} (t) - tan^{4} (t)

指数の規則を適用する:

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

sec^{4} (t) = (sec^{2} (t))^{2}

= (sec^{2} (t))^{2} - tan^{4} (t)

指数の規則を適用する:

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

tan^{4} (t) = (tan^{2} (t))^{2}

= (sec^{2} (t))^{2} - (tan^{2} (t))^{2}

= (sec^{2} (t))^{2} - (tan^{2} (t))^{2}

2乗の差の公式を適用する:

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

(sec^{2} (t))^{2} - (tan^{2} (t))^{2} = (sec^{2} (t) + tan^{2} (t)) (sec^{2} (t) - tan^{2} (t))

= (sec^{2} (t) + tan^{2} (t)) (sec^{2} (t) - tan^{2} (t))

因数

sec^{2} (t) - tan^{2} (t) : (sec (t) + tan (t)) (sec (t) - tan (t))

sec^{2} (t) - tan^{2} (t)

2乗の差の公式を適用する:

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

sec^{2} (t) - tan^{2} (t) = (sec (t) + tan (t)) (sec (t) - tan (t))

= (sec (t) + tan (t)) (sec (t) - tan (t))

= (sec^{2} (t) + tan^{2} (t)) (sec (t) + tan (t)) (sec (t) - tan (t))

= (sec (t) + tan (t)) (sec (t) - tan (t)) (sec^{2} (t) + tan^{2} (t))

三角関数の公式を使用して書き換える

(sec (t) + tan (t)) (sec (t) - tan (t)) (sec^{2} (t) + tan^{2} (t))

ピタゴラスの公式を使用する:

sec^{2} (x) = tan^{2} (x) + 1

= (sec (t) + tan (t)) (sec (t) - tan (t)) (tan^{2} (t) + 1 + tan^{2} (t))

簡素化

tan^{2} (t) + 1 + tan^{2} (t) : 2 tan^{2} (t) + 1

tan^{2} (t) + 1 + tan^{2} (t)

条件のようなグループ

= tan^{2} (t) + tan^{2} (t) + 1

類似した元を足す:

tan^{2} (t) + tan^{2} (t) = 2 tan^{2} (t)

= 2 tan^{2} (t) + 1

= (sec (t) + tan (t)) (sec (t) - tan (t)) (2 tan^{2} (t) + 1)

拡張

(sec (t) + tan (t)) (sec (t) - tan (t)) : sec^{2} (t) - tan^{2} (t)

(sec (t) + tan (t)) (sec (t) - tan (t))

2乗の差の公式を適用する:

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

a = sec (t), b = tan (t)

= sec^{2} (t) - tan^{2} (t)

= (sec^{2} (t) - tan^{2} (t)) (2 tan^{2} (t) + 1)

ピタゴラスの公式を使用する:

sec^{2} (t) = tan^{2} (t) + 1

sec^{2} (t) - tan^{2} (t) = 1

= 1 \cdot (2 tan^{2} (t) + 1)

簡素化

= 2 tan^{2} (t) + 1

= 2 tan^{2} (t) + 1

= 1 + 2 tan^{2} (t)

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(4θ)=cos^2(2θ)-sin^2(2θ)

p ro v e cos (4 θ) = cos^{2} (2 θ) - sin^{2} (2 θ)

証明する-30cos(3x)sin(3x)=-15sin(6x)

p ro v e - 30 cos (3 x) sin (3 x) = - 15 sin (6 x)

証明する csc^4(θ)-cot^4(θ)=2cot^2(θ)+1

p ro v e csc^{4} (θ) - cot^{4} (θ) = 2 cot^{2} (θ) + 1

証明する sec(b)cot(b)=csc(b)

p ro v e sec (b) cot (b) = csc (b)

証明する (1-cos(2θ))/(sin(2θ))=tan(θ)

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