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

Lösung

beweisen

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

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

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

Vereinfache

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

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

(cos (x) + sin (x))^{2} : cos^{2} (x) + 2 cos (x) sin (x) + sin^{2} (x)

Wende Formel für perfekte quadratische Gleichungen an:

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

a = cos (x), b = sin (x)

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

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

Addiere gleiche Elemente:

2 cos (x) sin (x) - 2 cos (x) sin (x) = 0

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

= 1

= 1

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

\Rightarrow Wahr

p ro v e (1 - sin (3 a)) (sin (3 a) + 1) = cos^{2} (3 a)

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

p ro v e sec (t) (csc (t) (tan (t) + cot (t))) = sec^{2} (t) + csc^{2} (t)

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

p ro v e cot (6 0^{\circ}) = \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )}