English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

Lösung

beweisen

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

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

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

Faktorisiere

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)

Schreibe

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

um:

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

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

Wende Exponentenregel an:

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

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

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

Wende Exponentenregel an:

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}

Wende Formel zur Differenz von zwei Quadraten an:

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))

Faktorisiere

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

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

Wende Formel zur Differenz von zwei Quadraten an:

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))

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

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

Vereinfache

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

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

Fasse gleiche Terme zusammen

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

Addiere gleiche Elemente:

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)

Multipliziere aus

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

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

Wende Formel zur Differenz von zwei Quadraten an:

(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)

Verwende die Pythagoreische Identität:

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

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

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

Vereinfache

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

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

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

Wir haben gezeigt, dass beide Seiten die gleiche Form annehmen können

\Rightarrow Wahr

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

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

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

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

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