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beweisen (1-cos^2(A))(1+cot^2(A))=1

Lösung

beweisen

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

Wahr

Schritte zur Lösung

(1 - cos^{2} (A)) (1 + cot^{2} (A)) = 1

Manipuliere die linke Seite

(1 - cos^{2} (A)) (1 + cot^{2} (A))

Drücke mit sin, cos aus

(1 + cot^{2} (A)) (1 - cos^{2} (A))

Verwende die grundlegende trigonometrische Identität:

cot (x) = \frac{cos ( x )}{sin ( x )}

= (1 + (\frac{cos ( A )}{sin ( A )})^{2}) (1 - cos^{2} (A))

Vereinfache

(1 + (\frac{cos ( A )}{sin ( A )})^{2}) (1 - cos^{2} (A)) : \frac{( sin ^{2} ( A ) + cos ^{2} ( A ) ) ( 1 - cos ^{2} ( A ) )}{sin ^{2} ( A )}

(1 + (\frac{cos ( A )}{sin ( A )})^{2}) (1 - cos^{2} (A))

Wende Exponentenregel an:

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

= (\frac{cos ^{2} ( A )}{sin ^{2} ( A )} + 1) (- cos^{2} (A) + 1)

Füge

1 + \frac{cos ^{2} ( A )}{sin ^{2} ( A )}

zusammen:

\frac{sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )}

1 + \frac{cos ^{2} ( A )}{sin ^{2} ( A )}

Wandle das Element in einen Bruch um:

1 = \frac{1 sin ^{2} ( A )}{sin ^{2} ( A )}

= \frac{1 \cdot sin ^{2} ( A )}{sin ^{2} ( A )} + \frac{cos ^{2} ( A )}{sin ^{2} ( A )}

Da die Nenner gleich sind, fasse die Brüche zusammen.:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )}

Multipliziere:

1 \cdot sin^{2} (A) = sin^{2} (A)

= \frac{sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )}

= \frac{sin ^{2} ( A ) + cos ^{2} ( A )}{sin ^{2} ( A )} (- cos^{2} (A) + 1)

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( sin ^{2} ( A ) + cos ^{2} ( A ) ) ( 1 - cos ^{2} ( A ) )}{sin ^{2} ( A )}

= \frac{( sin ^{2} ( A ) + cos ^{2} ( A ) ) ( 1 - cos ^{2} ( A ) )}{sin ^{2} ( A )}

= \frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + sin ^{2} ( A ) )}{sin ^{2} ( A )}

Umschreiben mit Hilfe von Trigonometrie-Identitäten

\frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + sin ^{2} ( A ) )}{sin ^{2} ( A )}

Verwende die Pythagoreische Identität:

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

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

= \frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + 1 - cos ^{2} ( A ) )}{1 - cos ^{2} ( A )}

\frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + 1 - cos ^{2} ( A ) )}{1 - cos ^{2} ( A )} = 1

\frac{( 1 - cos ^{2} ( A ) ) ( cos ^{2} ( A ) + 1 - cos ^{2} ( A ) )}{1 - cos ^{2} ( A )}

(1 - cos^{2} (A)) (cos^{2} (A) + 1 - cos^{2} (A)) = 1 - cos^{2} (A)

(1 - cos^{2} (A)) (cos^{2} (A) + 1 - cos^{2} (A))

Vereinfache

cos^{2} (A) + 1 - cos^{2} (A) : 1

cos^{2} (A) + 1 - cos^{2} (A)

Fasse gleiche Terme zusammen

= cos^{2} (A) - cos^{2} (A) + 1

Addiere gleiche Elemente:

cos^{2} (A) - cos^{2} (A) = 0

= 1

= 1 \cdot (- cos^{2} (A) + 1)

Multipliziere:

(1 - cos^{2} (A)) \cdot 1 = (1 - cos^{2} (A))

= (1 - cos^{2} (A))

Entferne die Klammern:

(a) = a

= 1 - cos^{2} (A)

= \frac{1 - cos ^{2} ( A )}{1 - cos ^{2} ( A )}

Wende Regel an

\frac{a}{a} = 1

= 1

= 1

= 1

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

\Rightarrow Wahr

p ro v e 4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)

p ro v e \frac{sin ^{2} ( 2 x )}{4} = sin^{2} (x) cos^{2} (x)

p ro v e \frac{sec ( θ )}{tan ( θ ) + cot ( θ )} = sin (θ)

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

p ro v e 7 cos^{2} (x) + 5 sin^{2} (x) = 6 + cos (2 x)