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beweisen coth^2(x)-1=csch^2(x)

Lösung

beweisen

coth^{2} (x) - 1 = csch^{2} (x)

Wahr

Schritte zur Lösung

coth^{2} (x) - 1 = csch^{2} (x)

Manipuliere die linke Seite

coth^{2} (x) - 1

Umschreiben mit Hilfe von Trigonometrie-Identitäten

coth^{2} (x) - 1

Hyperbolische Identität anwenden:

coth (x) = \frac{cosh ( x )}{sinh ( x )}

= (\frac{cosh ( x )}{sinh ( x )})^{2} - 1

Schreibe

1

um:

\frac{sinh ^{2} ( x )}{sinh ^{2} ( x )}

= (\frac{cosh ( x )}{sinh ( x )})^{2} - \frac{sinh ^{2} ( x )}{sinh ^{2} ( x )}

Vereinfache

(\frac{cosh ( x )}{sinh ( x )})^{2} - \frac{sinh ^{2} ( x )}{sinh ^{2} ( x )} : \frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

(\frac{cosh ( x )}{sinh ( x )})^{2} - \frac{sinh ^{2} ( x )}{sinh ^{2} ( x )}

Wende Regel an

\frac{a}{a} = 1

\frac{sinh ^{2} ( x )}{sinh ^{2} ( x )} = 1

= (\frac{cosh ( x )}{sinh ( x )})^{2} - 1

Wende Exponentenregel an:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{cosh ^{2} ( x )}{sinh ^{2} ( x )} - 1

Wandle das Element in einen Bruch um:

1 = \frac{1 sinh ^{2} ( x )}{sinh ^{2} ( x )}

= \frac{cosh ^{2} ( x )}{sinh ^{2} ( x )} - \frac{1 sinh ^{2} ( x )}{sinh ^{2} ( x )}

Da die Nenner gleich sind, fasse die Brüche zusammen.:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cosh ^{2} ( x ) - 1 sinh ^{2} ( x )}{sinh ^{2} ( x )}

Multipliziere:

1 \cdot sinh^{2} (x) = sinh^{2} (x)

= \frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

= \frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

= \frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

Umschreiben mit Hilfe von Trigonometrie-Identitäten

\frac{cosh ^{2} ( x ) - sinh ^{2} ( x )}{sinh ^{2} ( x )}

Hyperbolische Identität anwenden:

cosh^{2} (x) - sinh^{2} (x) = 1

= \frac{1}{sinh ^{2} ( x )}

Hyperbolische Identität anwenden:

sinh (x) = \frac{1}{csch ( x )}

= \frac{1}{( \frac{1}{csch ( x )} ) ^{2}}

Vereinfache

\frac{1}{( \frac{1}{csch ( x )} ) ^{2}} : csch^{2} (x)

\frac{1}{( \frac{1}{csch ( x )} ) ^{2}}

(\frac{1}{csch ( x )})^{2} = \frac{1}{csch ^{2} ( x )}

(\frac{1}{csch ( x )})^{2}

Wende Exponentenregel an:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{csch ^{2} ( x )}

Wende Regel an

1^{a} = 1

1^{2} = 1

= \frac{1}{csch ^{2} ( x )}

= \frac{1}{\frac{1}{csch ^{2} ( x )}}

Wende Bruchregel an:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csch ^{2} ( x )}{1}

Wende Regel an

\frac{a}{1} = a

= csch^{2} (x)

= csch^{2} (x)

= csch^{2} (x)

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

\Rightarrow Wahr

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

p ro v e cot (\frac{π}{2} - x) = tan (x)

p ro v e cot^{2} (x) - csc^{2} (x) = - 1

p ro v e cot (a) sec (a) = csc (a)

p ro v e tan (\frac{x}{2}) + cot (\frac{x}{2}) = 2 csc (x)