faktorisieren cos^6(θ)+sin^6(θ)

Lösung

faktorisieren

cos^{6} (θ) + sin^{6} (θ)

(cos^{2} (θ) + sin^{2} (θ)) (cos^{4} (θ) - sin^{2} (θ) cos^{2} (θ) + sin^{4} (θ))

Schritte zur Lösung

cos^{6} (θ) + sin^{6} (θ)

Schreibe

cos^{6} (θ) + sin^{6} (θ)

um:

(cos^{2} (θ))^{3} + (sin^{2} (θ))^{3}

= (cos^{2} (θ))^{3} + (sin^{2} (θ))^{3}

Wende Formel zur Summe von dritten Potenzen an:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(cos^{2} (θ))^{3} + (sin^{2} (θ))^{3} = (cos^{2} (θ) + sin^{2} (θ)) (cos^{4} (θ) - sin^{2} (θ) cos^{2} (θ) + sin^{4} (θ))

= (sin^{2} (θ) + cos^{2} (θ)) (sin^{4} (θ) - sin^{2} (θ) cos^{2} (θ) + cos^{4} (θ))

f a c t or (- 4 + 43 i)^{20}

f a c t or (x) - 1

\frac{d}{d t} ((x) (sin (k t)))

f a c t or sin^{2} (x) - 2 sin (x) + 1

f a c t or x (3 cos (2 x))