Arccos of neg.number

[ittiandro]

Hi!

I hit a wall on the following trig formula:

Z=arccosin (sin45xsin(-23.50) + cos45Xcos(-23.50)).

After working out the parentheses contents, I get :

Z=arcos-32.50

why does my calculator return an error message for arcos(-32.50), i.e. a neg. number??

For your information, Z is the the Zenith angle, i.e. the angle between the sun's observed position in the sky ( depending on the latitude, the time of the day and the season) and its would-be Zenith position.

Z , in turn, allows to calculate the insolation value ( I) for that latitude , season and time of the day, according to the formula I=SXcosZ, where S is the amount of heat energy received by the Earth ( insolation) when the sun is perpendicular ( 1000w/Sqmt/sec)

In the above formula I have already plugged in those values: the value 45 deg.( sin45) is the latitude I chose , while -23.50 deg. is the solar declination angle corresponding to the winter solstice, as opposed to + 23.50 for the summer solstice.

Thank you for your help

Franco

calculator gives me an error message for the negative value of -23.50 deg., which is the correct solar declination angle on the winter solstice, as compared with + 23 50 deg.value for the summer solstice. I have no problems in detemining Z for + 23.50 deg. Why do I get an error message for a negative value such as -23.50 ?

[Lord Orwell]

because that function only works on positive degrees. try converting your impossibly negative degree by subtracting it from 360 and see if it helps.

[ittiandro]

Thank you, Lord Orwell.

I did as you said and I do get an arccos value. At least there is no more error message.
However, the result I get ( the insolation value at 45 deg. latitude on the winter solstice ) does not seem to be consistent with the insolation values I get for other latitudes and the other seasonal cycles ( summer solstice and equinoxes). Or may be the maths are right and it is me that has overlooked something.

Since I always try to understand, rather than using maths blindly, I'd love to know, though, why the problem could be solved by subtracting -32.50 from 360 deg.( 360-32.50 ) Or is it 360-(-32.50)? I did the first. because otherwise I would have got more than 360 deg, which does not make sense. Am I right?


Thank you again

Franco

[Lord Orwell]

try this page
http://oakroadsystems.com/twt/inverse.htm
this math is beyond me. I didn't take trig to that level, but there's something in there at arccos a negative number.
Also if you still have your calculator's manual, try looking that function up. There may be a special way to enter it with negative numbers.

[jemidiah]

The domain (i.e. set of valid inputs) of the arccos function is all numbers from -1 to 1, being the range (i.e. set of valid outputs) of the cos function, its inverse. Negative values are indeed allowed so long as they're between -1 and 0. I don't know how you got "sin45xsin(-23.50) + cos45Xcos(-23.50)" (which I'm interpreting as "sin 45 deg * sin -23.5 deg + cos 45 deg * cos -23.5 deg with * meaning multiplication) equal to -32.5. I get 0.366501227. arccos(360 - 32.5) could not evaluate without an error under the standard definitions, so I don't know what you did to get an answer.

To be clear, the input to the inverse trig functions like arccos is *not* an angle measurement, so is not in degrees or radians.

I'm interpreting "arccosin" as the usual arccos or acos.