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why is arccos of -3.225 invalid?

 Apr 23, 2014

Best Answer 

 #3
avatar+118703 
+5

Ignore the minus sign for a moment

Think of a right angled triangle

cos(theta)=adjacent / hypotenuse

The hypotenuse is the longest side so cos(theta) must be between 0 and 1 (for a right triangle 0 and 1 are not included)

When you extend this idea to include all values of cos then it can be between -1 and 1

 Apr 24, 2014
 #1
avatar+33654 
+5

This would mean the angle whose cosine is -3.2257; but the limits of cosine lie between +1 and -1 (inclusive).

 Apr 23, 2014
 #2
avatar+130466 
+5

why is arccos of -3.225 invalid?

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In a simple manner, we might say that the the arccos function is basically asking, "What angle has a cosine value of -3.225??"

But the cosine value of an angle can only "range" from -1 to 1, inclusive. So no angle exists with a cosine value of -3.225. That's why you get an "invalid" or "error" message on your calculator. Also, look at the graph of the cosine curve........you'll see that the "crests" are never "greater" than "1," and the "troughs" are never "less" than "-1." 

 Apr 23, 2014
 #3
avatar+118703 
+5
Best Answer

Ignore the minus sign for a moment

Think of a right angled triangle

cos(theta)=adjacent / hypotenuse

The hypotenuse is the longest side so cos(theta) must be between 0 and 1 (for a right triangle 0 and 1 are not included)

When you extend this idea to include all values of cos then it can be between -1 and 1

Melody Apr 24, 2014

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