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May 31st, 2015, 10:01 AM   #1
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Maximum/Minimum value of trigonometry function


This is a common CIE A levels P1 paper question, but I seem to have forgotten how to do it:

Find the maximum value and the minimum value of


8/[3-sin x]

at the least positive value of x at which it occurs.

I understand doing 11-5cos(x/2-45) as being


and so on... but how do I do this if there is a fraction?

Any help would be appreciated, thank you in advance!
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May 31st, 2015, 10:19 AM   #2
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As both denominators are always positive you don't have to consider negative values. With
the denominator and numerator positive, a fraction with a constant numerator is at a maximum
where the denominator is smallest and at a minimum where the denominator is largest.
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May 31st, 2015, 11:40 AM   #3
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Hmm... sorry, I still don't understand that. How would you solve the above 2 questions though? Thank you
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