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Simplify cofunction expression

  • Thread starter synergix
  • Start date
  • #1
178
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Homework Statement


csc(pi/2-x)/cos(x+pi/2) + cot(pi/2-x)

The Attempt at a Solution



1/cosx/-sinx + sinx/cosx

-1/sinxcosx+ sinx/cosx(sinx/sinx)

(sin2x - 1) / sinxcosx

-cos2x/sinxcosx

-cosx/sinx

-cotx

is this right the answer key says it is cosx but that could be wrong I have done this a couple times and gotten the same answer
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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I don't think it can be the same as cosx because if you put x=0, it doesn't work.
 
  • #3
52
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csc(pi/2-x)/cos(x+pi/2) + cot(pi/2-x)
1/cosx/-sinx + sinx/cosx
I'm assuming you mean (1/cosx)/-sinx in the first fraction. Anyway, this is not correct.


01
 
  • #4
178
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(1)
cofunction%20identiies%20csc%20sec.gif


(2)cos(x+pi/2) = - sinx

(1)/(2)=

secx/-sinx=

(1/cosx)/-sinx

what is wrong about that?
 
  • #5
52
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Oops, sorry about that. I misread the problem. I redid the problem and now I'm getting -cot x. You sure you copied the problem correctly?


01
 
  • #6
178
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Well the way it is written is the csc cofunction is directly above the cos cofunction and then added to the cot cofunction. There isn't actually a line between the two. I am sure i am meant to divide the two but is there anything else that could mean. FYI my instructor is very smart but he is also somewhat absent minded and I am pretty sure he put together these practice assignments himself he could have made a mistake it wouldn't be the first time.
 
  • #7
jhae2.718
Gold Member
1,161
20
Assuming the original problem was copied correctly, I would say that you are correct; I get -cot(x) when I solve the problem. In addition I graphed the two curves as a check and they are the same...
 

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