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beweisen 1/(sec^2(x))+1/(csc^2(x))=1

Lösung

beweisen

\frac{1}{sec ^{2} ( x )} + \frac{1}{csc ^{2} ( x )} = 1

Wahr

Schritte zur Lösung

\frac{1}{sec ^{2} ( x )} + \frac{1}{csc ^{2} ( x )} = 1

Manipuliere die linke Seite

\frac{1}{sec ^{2} ( x )} + \frac{1}{csc ^{2} ( x )}

Drücke mit sin, cos aus

\frac{1}{csc ^{2} ( x )} + \frac{1}{sec ^{2} ( x )}

Verwende die grundlegende trigonometrische Identität:

csc (x) = \frac{1}{sin ( x )}

= \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{sec ^{2} ( x )}

Verwende die grundlegende trigonometrische Identität:

sec (x) = \frac{1}{cos ( x )}

= \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

Vereinfache

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}} : sin^{2} (x) + cos^{2} (x)

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} + \frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} = sin^{2} (x)

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

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

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

Wende Exponentenregel an:

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

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

Wende Regel an

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{s i n ^{2} ( x )}}

Wende Bruchregel an:

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

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

Wende Regel an

\frac{a}{1} = a

= sin^{2} (x)

\frac{1}{( \frac{1}{c o s ( x )} ) ^{2}} = cos^{2} (x)

\frac{1}{( \frac{1}{c o s ( x )} ) ^{2}}

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

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

Wende Exponentenregel an:

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

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

Wende Regel an

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{c o s ^{2} ( x )}}

Wende Bruchregel an:

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

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

Wende Regel an

\frac{a}{1} = a

= cos^{2} (x)

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

= sin^{2} (x) + cos^{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 + tan^{2} (x)) cot^{2} (x) = csc^{2} (x)

p ro v e 1 - tan^{2} (\frac{θ}{2}) = \frac{2 cos ( θ )}{1 + cos ( θ )}

p ro v e sin (x + y) sec (x) sec (y) = tan (x) + tan (y)

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

p ro v e tan (2 x) - 2 tan (2 x) sin^{2} (x) = sin (2 x)