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beweisen csc^4(θ)-cot^4(θ)=2cot^2(θ)+1

Lösung

beweisen

csc^{4} (θ) - cot^{4} (θ) = 2 cot^{2} (θ) + 1

Wahr

Schritte zur Lösung

csc^{4} (θ) - cot^{4} (θ) = 2 cot^{2} (θ) + 1

Manipuliere die linke Seite

csc^{4} (θ) - cot^{4} (θ)

Faktorisiere

- cot^{4} (θ) + csc^{4} (θ) : (csc^{2} (θ) + cot^{2} (θ)) (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

- cot^{4} (θ) + csc^{4} (θ)

Schreibe

csc^{4} (θ) - cot^{4} (θ)

um:

(csc^{2} (θ))^{2} - (cot^{2} (θ))^{2}

csc^{4} (θ) - cot^{4} (θ)

Wende Exponentenregel an:

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

cot^{4} (θ) = (cot^{2} (θ))^{2}

= csc^{4} (θ) - (cot^{2} (θ))^{2}

Wende Exponentenregel an:

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

csc^{4} (θ) = (csc^{2} (θ))^{2}

= (csc^{2} (θ))^{2} - (cot^{2} (θ))^{2}

= (csc^{2} (θ))^{2} - (cot^{2} (θ))^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

(csc^{2} (θ))^{2} - (cot^{2} (θ))^{2} = (csc^{2} (θ) + cot^{2} (θ)) (csc^{2} (θ) - cot^{2} (θ))

= (csc^{2} (θ) + cot^{2} (θ)) (csc^{2} (θ) - cot^{2} (θ))

Faktorisiere

csc^{2} (θ) - cot^{2} (θ) : (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

csc^{2} (θ) - cot^{2} (θ)

Wende Formel zur Differenz von zwei Quadraten an:

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

csc^{2} (θ) - cot^{2} (θ) = (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

= (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

= (csc^{2} (θ) + cot^{2} (θ)) (csc (θ) + cot (θ)) (csc (θ) - cot (θ))

= (- cot (θ) + csc (θ)) (cot (θ) + csc (θ)) (cot^{2} (θ) + csc^{2} (θ))

Umschreiben mit Hilfe von Trigonometrie-Identitäten

(- cot (θ) + csc (θ)) (cot (θ) + csc (θ)) (cot^{2} (θ) + csc^{2} (θ))

Multipliziere aus

(csc (θ) + cot (θ)) (csc (θ) - cot (θ)) : csc^{2} (θ) - cot^{2} (θ)

(csc (θ) + cot (θ)) (csc (θ) - cot (θ))

Wende Formel zur Differenz von zwei Quadraten an:

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

a = csc (θ), b = cot (θ)

= csc^{2} (θ) - cot^{2} (θ)

= (csc^{2} (θ) - cot^{2} (θ)) (cot^{2} (θ) + csc^{2} (θ))

Verwende die Pythagoreische Identität:

csc^{2} (θ) = 1 + cot^{2} (θ)

csc^{2} (θ) - cot^{2} (θ) = 1

= 1 \cdot (cot^{2} (θ) + csc^{2} (θ))

Vereinfache

= cot^{2} (θ) + csc^{2} (θ)

Verwende die Pythagoreische Identität:

csc^{2} (x) = 1 + cot^{2} (x)

= cot^{2} (θ) + 1 + cot^{2} (θ)

Vereinfache

cot^{2} (θ) + 1 + cot^{2} (θ) : 2 cot^{2} (θ) + 1

cot^{2} (θ) + 1 + cot^{2} (θ)

Fasse gleiche Terme zusammen

= cot^{2} (θ) + cot^{2} (θ) + 1

Addiere gleiche Elemente:

cot^{2} (θ) + cot^{2} (θ) = 2 cot^{2} (θ)

= 2 cot^{2} (θ) + 1

= 2 cot^{2} (θ) + 1

= 2 cot^{2} (θ) + 1

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

\Rightarrow Wahr

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

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

p ro v e \frac{cos ( 3 x )}{sin ( x )} + \frac{sin ( 3 x )}{cos ( x )} = 2 cot (2 x)

p ro v e sin (A + B) = sin (A) + sin (B)

p ro v e \frac{cot ^{2} ( B ) - cos ^{2} ( B )}{csc ^{2} ( B ) - 1} = cos^{2} (B)