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

Lösung

beweisen

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

Wahr

Schritte zur Lösung

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

Manipuliere die linke Seite

\frac{1}{csc ^{2} ( x )} + \frac{1}{sec ^{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 cot (\frac{5 π}{12}) = 2 - 3

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p ro v e \frac{cos ^{3} ( t )}{cot ^{2} ( t )} = sin^{2} (t) cos (t)

p ro v e cos (a + b) cos (a - b) = cos^{2} (a) - cos^{2} (b)

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