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

Lösung

beweisen

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

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

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

Vereinfache

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

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

Faktorisiere

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

- 1 + csc^{4} (x)

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

Streiche die gemeinsamen Faktoren:

csc^{2} (x) + 1

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Multipliziere aus

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

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

Wende Formel zur Differenz von zwei Quadraten an:

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

a = csc (x), b = 1

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

Wende Regel an

1^{a} = 1

1^{2} = 1

= csc^{2} (x) - 1

= csc^{2} (x) - 1

Verwende die Pythagoreische Identität:

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

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

= cot^{2} (x)

= cot^{2} (x)

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

\Rightarrow Wahr

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

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

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

p ro v e tan^{2} (y) (csc^{2} (y) - 1) = 1

p ro v e cos (2 π + x) = cos (x)