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Warum ist cos^-1(1.18) nicht zu errechnen?

 14.12.2014

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 #1
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$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1.18}}\right)} = \underset{{\tiny{\text{Error: invalid argument}}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1.18}}\right)}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0}}\right)} = {\mathtt{90^{\circ}}}$$      ;    $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.5}}\right)} = {\mathtt{60^{\circ}}}$$      ;   $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1}}\right)} = {\mathtt{0^{\circ}}}$$

 

cos^-1   (arcuscos)   für Werte > 1 nicht definiert !

Gruß radix !

 14.12.2014
 #1
avatar+14538 
+5
Beste Antwort

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1.18}}\right)} = \underset{{\tiny{\text{Error: invalid argument}}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1.18}}\right)}}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0}}\right)} = {\mathtt{90^{\circ}}}$$      ;    $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.5}}\right)} = {\mathtt{60^{\circ}}}$$      ;   $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{1}}\right)} = {\mathtt{0^{\circ}}}$$

 

cos^-1   (arcuscos)   für Werte > 1 nicht definiert !

Gruß radix !

radix 14.12.2014
 #2
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Danke (:

 14.12.2014

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