beweisen (1+cos(2θ))(1-cos(2θ))=sin^2(2θ)

Lösung

beweisen

(1 + cos (2 θ)) (1 - cos (2 θ)) = sin^{2} (2 θ)

Wahr

Schritte zur Lösung

(1 + cos (2 θ)) (1 - cos (2 θ)) = sin^{2} (2 θ)

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

(1 + cos (x)) (1 - cos (x))

Multipliziere aus

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

(1 + cos (x)) (1 - cos (x))

Wende Formel zur Differenz von zwei Quadraten an:

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

a = 1, b = cos (x)

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

Wende Regel an

1^{a} = 1

1^{2} = 1

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

Verwende die Pythagoreische Identität:

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

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

= sin^{2} (x)

\Rightarrow Wahr

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

p ro v e cos (27 0^{\circ} - θ) = - sin (θ)

p ro v e cot (2 a) = \frac{1}{2} (cot (a) - tan (a))

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

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