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

Lösung

beweisen

\frac{csc ^{4} ( x ) - 1}{cot ^{2} ( x )} = csc^{2} (x) + 1

Wahr

Schritte zur Lösung

\frac{csc ^{4} ( x ) - 1}{cot ^{2} ( x )} = csc^{2} (x) + 1

Manipuliere die linke Seite

\frac{csc ^{4} ( x ) - 1}{cot ^{2} ( x )}

Umschreiben mit Hilfe von Trigonometrie-Identitäten

\frac{csc ^{4} ( x ) - 1}{cot ^{2} ( x )}

Verwende die Pythagoreische Identität:

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

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

= \frac{csc ^{4} ( x ) - 1}{csc ^{2} ( x ) - 1}

Vereinfache

\frac{csc ^{4} ( x ) - 1}{csc ^{2} ( x ) - 1} : csc^{2} (x) + 1

\frac{csc ^{4} ( x ) - 1}{csc ^{2} ( x ) - 1}

Faktorisiere

csc^{4} (x) - 1 : (csc^{2} (x) + 1) (csc (x) + 1) (csc (x) - 1)

csc^{4} (x) - 1

Schreibe

csc^{4} (x) - 1

um:

(csc^{2} (x))^{2} - 1^{2}

csc^{4} (x) - 1

Schreibe

1

um:

1^{2}

= csc^{4} (x) - 1^{2}

Wende Exponentenregel an:

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

csc^{4} (x) = (csc^{2} (x))^{2}

= (csc^{2} (x))^{2} - 1^{2}

= (csc^{2} (x))^{2} - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

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

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

Faktorisiere

csc^{2} (x) - 1 : (csc (x) + 1) (csc (x) - 1)

csc^{2} (x) - 1

Schreibe

1

um:

1^{2}

= csc^{2} (x) - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (csc (x) + 1) (csc (x) - 1)

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

= \frac{( csc ^{2} ( x ) + 1 ) ( csc ( x ) + 1 ) ( csc ( x ) - 1 )}{csc ^{2} ( x ) - 1}

Faktorisiere

csc^{2} (x) - 1 : (csc (x) + 1) (csc (x) - 1)

csc^{2} (x) - 1

Schreibe

1

um:

1^{2}

= csc^{2} (x) - 1^{2}

Wende Formel zur Differenz von zwei Quadraten an:

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

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

= (csc (x) + 1) (csc (x) - 1)

= \frac{( csc ^{2} ( x ) + 1 ) ( csc ( x ) + 1 ) ( csc ( x ) - 1 )}{( csc ( x ) + 1 ) ( csc ( x ) - 1 )}

Streiche

\frac{( csc ^{2} ( x ) + 1 ) ( csc ( x ) + 1 ) ( csc ( x ) - 1 )}{( csc ( x ) + 1 ) ( csc ( x ) - 1 )} : csc^{2} (x) + 1

\frac{( csc ^{2} ( x ) + 1 ) ( csc ( x ) + 1 ) ( csc ( x ) - 1 )}{( csc ( x ) + 1 ) ( csc ( x ) - 1 )}

Streiche die gemeinsamen Faktoren:

csc (x) + 1

= \frac{( csc ^{2} ( x ) + 1 ) ( csc ( x ) - 1 )}{csc ( x ) - 1}

Streiche die gemeinsamen Faktoren:

csc (x) - 1

= csc^{2} (x) + 1

= csc^{2} (x) + 1

= csc^{2} (x) + 1

= csc^{2} (x) + 1

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

\Rightarrow Wahr

p ro v e sin (θ) + cos (θ) = 2 sin (θ + 4 5^{\circ})

p ro v e \frac{1 - tan ( θ )}{1 + tan ( θ )} = \frac{cot ( θ ) - 1}{cot ( θ ) + 1}

p ro v e cot (θ) sin (θ) cos (θ) = 1 - sin^{2} (θ)

p ro v e csc (θ) = \frac{1}{sin ( θ )}

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