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

Lösung

prove

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Verwende die Pythagoreische Identität:

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

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

Vereinfache

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

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

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

Wende Formel für perfekte quadratische Gleichungen an:

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

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

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

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

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

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

Setze Klammern

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

Wende Minus-Plus Regeln an

- (- a) = a, - (a) = - a

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

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

Vereinfache

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

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

Addiere gleiche Elemente:

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

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

Addiere gleiche Elemente:

sin^{2} (x) - sin^{2} (x) = 0

= 2 sin (x) cos (x)

= 2 sin (x) cos (x)

= 2 sin (x) cos (x)

Verwende die Doppelwinkelidentität:

2 sin (x) cos (x) = sin (2 x)

= sin (2 x)

= sin (2 x)

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

\Rightarrow Wahr