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beweisen cot(x)-csc^2(x)cot(x)=-cot^3(x)

Lösung

beweisen

cot (x) - csc^{2} (x) cot (x) = - cot^{3} (x)

Wahr

Schritte zur Lösung

cot (x) - csc^{2} (x) cot (x) = - cot^{3} (x)

Manipuliere die linke Seite

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

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

Vereinfache

cot (x) - (1 + cot^{2} (x)) cot (x) : - cot^{3} (x)

cot (x) - (1 + cot^{2} (x)) cot (x)

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

Multipliziere aus

- cot (x) (1 + cot^{2} (x)) : - cot (x) - cot^{3} (x)

- cot (x) (1 + cot^{2} (x))

Wende das Distributivgesetz an:

a (b + c) = ab + a c

a = - cot (x), b = 1, c = cot^{2} (x)

= - cot (x) \cdot 1 + (- cot (x)) cot^{2} (x)

Wende Minus-Plus Regeln an

+ (- a) = - a

= - 1 \cdot cot (x) - cot^{2} (x) cot (x)

Vereinfache

- 1 \cdot cot (x) - cot^{2} (x) cot (x) : - cot (x) - cot^{3} (x)

- 1 \cdot cot (x) - cot^{2} (x) cot (x)

1 \cdot cot (x) = cot (x)

1 \cdot cot (x)

Multipliziere:

1 \cdot cot (x) = cot (x)

= cot (x)

cot^{2} (x) cot (x) = cot^{3} (x)

cot^{2} (x) cot (x)

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

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

= cot^{2 + 1} (x)

Addiere die Zahlen:

2 + 1 = 3

= cot^{3} (x)

= - cot (x) - cot^{3} (x)

= - cot (x) - cot^{3} (x)

= cot (x) - cot (x) - cot^{3} (x)

Addiere gleiche Elemente:

cot (x) - cot (x) = 0

= - cot^{3} (x)

= - cot^{3} (x)

= - cot^{3} (x)

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

\Rightarrow Wahr

p ro v e cot (x - \frac{π}{2}) = - tan (x)

p ro v e \frac{1 + tan ^{2} ( θ )}{1 - tan ^{2} ( θ )} = sec (2 θ)

p ro v e cos^{2} (3 x) - sin^{2} (3 x) = cos (6 x)

p ro v e \frac{cos ( t )}{1 - sin ( t )} = sec (t) + tan (t)

p ro v e sin (2 x) = 2 \frac{cos ( x )}{csc ( x )}