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beweisen tan^8(x)=tan^6(x)sec^2(x)-tan^6(x)

Lösung

beweisen

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Wahr

Schritte zur Lösung

tan^{8} (x) = tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Manipuliere die rechte Seite

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Umschreiben mit Hilfe von Trigonometrie-Identitäten

tan^{6} (x) sec^{2} (x) - tan^{6} (x)

Verwende die Pythagoreische Identität:

sec^{2} (x) = tan^{2} (x) + 1

= tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

Vereinfache

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x) : tan^{8} (x)

tan^{6} (x) (tan^{2} (x) + 1) - tan^{6} (x)

Multipliziere aus

tan^{6} (x) (tan^{2} (x) + 1) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) (tan^{2} (x) + 1)

Wende das Distributivgesetz an:

a (b + c) = ab + a c

a = tan^{6} (x), b = tan^{2} (x), c = 1

= tan^{6} (x) tan^{2} (x) + tan^{6} (x) \cdot 1

= tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

Vereinfache

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x) : tan^{8} (x) + tan^{6} (x)

tan^{6} (x) tan^{2} (x) + 1 \cdot tan^{6} (x)

tan^{6} (x) tan^{2} (x) = tan^{8} (x)

tan^{6} (x) tan^{2} (x)

Wende Exponentenregel an:

a^{b} \cdot a^{c} = a^{b + c}

tan^{6} (x) tan^{2} (x) = tan^{6 + 2} (x)

= tan^{6 + 2} (x)

Addiere die Zahlen:

6 + 2 = 8

= tan^{8} (x)

1 \cdot tan^{6} (x) = tan^{6} (x)

1 \cdot tan^{6} (x)

Multipliziere:

1 \cdot tan^{6} (x) = tan^{6} (x)

= tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x)

= tan^{8} (x) + tan^{6} (x) - tan^{6} (x)

Addiere gleiche Elemente:

tan^{6} (x) - tan^{6} (x) = 0

= tan^{8} (x)

= tan^{8} (x)

= tan^{8} (x)

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

\Rightarrow Wahr

p ro v e 2 (sin (2 y) cos (2 y)) = sin (4 y)

p ro v e \frac{( tan ^{2} ( t ) )}{sec ( t )} = sec (t) - cos (t)

p ro v e - sin (- 1) = sin (1)

p ro v e sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

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