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beweisen cos^4(β)-sin^4(β)=cos(2β)

Lösung

beweisen

cos^{4} (β) - sin^{4} (β) = cos (2 β)

Wahr

Schritte zur Lösung

cos^{4} (β) - sin^{4} (β) = cos (2 β)

Manipuliere die linke Seite

cos^{4} (β) - sin^{4} (β)

Faktorisiere

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

cos^{4} (β) - sin^{4} (β)

Schreibe

cos^{4} (β) - sin^{4} (β)

um:

(cos^{2} (β))^{2} - (sin^{2} (β))^{2}

cos^{4} (β) - sin^{4} (β)

Wende Exponentenregel an:

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

sin^{4} (β) = (sin^{2} (β))^{2}

= cos^{4} (β) - (sin^{2} (β))^{2}

Wende Exponentenregel an:

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

cos^{4} (β) = (cos^{2} (β))^{2}

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

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

Wende Formel zur Differenz von zwei Quadraten an:

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

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

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

Faktorisiere

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

cos^{2} (β) - sin^{2} (β)

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (cos (β) + sin (β)) (cos (β) - sin (β))

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

= (cos (β) + sin (β)) (cos (β) - sin (β)) \cdot 1

Vereinfache

= (cos (β) + sin (β)) (cos (β) - sin (β))

Multipliziere aus

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

(cos (β) + sin (β)) (cos (β) - sin (β))

Wende Formel zur Differenz von zwei Quadraten an:

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

a = cos (β), b = sin (β)

= cos^{2} (β) - sin^{2} (β)

= cos^{2} (β) - sin^{2} (β)

Verwende die Doppelwinkelidentität:

cos^{2} (β) - sin^{2} (β) = cos (2 β)

= cos (2 β)

= cos (2 β)

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

\Rightarrow Wahr

p ro v e cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

p ro v e tan (\frac{π}{4} + x) = \frac{1 + sin ( 2 x )}{cos ( 2 x )}

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

p ro v e sin (x) = 1 - cos (x)

p ro v e \frac{sec ( 2 x ) + 1}{tan ( 2 x )} = cot (x)