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人気のある三角関数 >

証明する tan^4(k)-sec^4(k)=1-2sec^2(k)

解

証明する

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

解

真

解答ステップ

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

左側を操作する

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

因数

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

- sec^{4} (k) + tan^{4} (k)

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

を書き換え

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

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

指数の規則を適用する:

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

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

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

指数の規則を適用する:

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

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

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

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

2乗の差の公式を適用する:

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

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

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

因数

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

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

2乗の差の公式を適用する:

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

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

= (tan (k) + sec (k)) (tan (k) - sec (k))

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

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

類似した元を足す:

sec^{2} (k) + sec^{2} (k) = 2 sec^{2} (k)

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

拡張

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

(tan (k) + sec (k)) (tan (k) - sec (k))

2乗の差の公式を適用する:

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

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

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

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

ピタゴラスの公式を使用する:

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

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

= (- 1 + 2 sec^{2} (k)) (- 1)

簡素化

(- 1 + 2 sec^{2} (k)) (- 1) : 1 - 2 sec^{2} (k)

(- 1 + 2 sec^{2} (k)) (- 1)

括弧を削除する:

(- a) = - a

= - (- 1 + 2 sec^{2} (k)) \cdot 1

乗算:

(- 1 + 2 sec^{2} (k)) \cdot 1 = (- 1 + 2 sec^{2} (k))

= - (2 sec^{2} (k) - 1)

括弧を分配する

= - (- 1) - (2 sec^{2} (k))

マイナス・プラスの規則を適用する

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

= 1 - 2 sec^{2} (k)

= 1 - 2 sec^{2} (k)

= 1 - 2 sec^{2} (k)

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 1-tan^4(x)=2sec^2(x)-sec^4(x)

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

証明する sin(pi)=sin(-pi)

p ro v e sin (π) = sin (- π)

証明する csc(t)cot(t)=(1+cot^2(t))/(sec(t))

p ro v e csc (t) cot (t) = \frac{1 + cot ^{2} ( t )}{sec ( t )}

証明する csc(a)sec(a)=2csc(2a)

p ro v e csc (a) sec (a) = 2 csc (2 a)

証明する tan^2(2x)+1=sec^2(2x)

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