Is any one smart in Calculus.
#16
Originally posted by Makaveli
Yeah, why did I write pie too? Redrims, you're a bad influence
So were the answers right???
The whole forum is waiting on the answers
Yeah, why did I write pie too? Redrims, you're a bad influence
So were the answers right???
The whole forum is waiting on the answers
#17
Re: Is any one smart in Calculus.
Hey, I know I am a little late for this but, here goes anyways, the question isnt terribly hard, you just have to be logical.
Basically they have given you enough information to determine what t is.
tant=4/5
therefore t must be 38.65980825...
but wait, if we look at the tan graph we also see that 4/5 or 0.8 can also occur at 38.65 plus pi or 3pi or 5pi etc.
They have also told us that cost must be less than zero, therefore if we look at the cos graph we see that cos can only be zero between 90 and 270, or 450 and 630 etc. This means that t must fall on the tan graph in those same areas as allowed on the cos chart.
So we see that t can only meet the criteria when it equals 38.65+180 or 540 etc. written out this is
t=(38.65+pi+2Xpi)
(I am giving the very general case in that equation)
So the first question a
cost=cos(38.65+pi+0)=-0.78, which meets the criteria of cost<0
second question
sin(-t) = sin -(38.65+pi)=0.624
third
cot(t+pi) = 38.65+pi+pi =1.25
fourth
sec(t+6pi)=sec(38.65+pi+6pi) or 1/cos(...)=1.28
of course you could just realize too that the 6pi is just the 2xpi term in my equation above and get rid of it as it doesnt affect the final answer. This goes for the third question too.
hope this helps.
Stocker.
If you need actual help in calculus, just ask, I did 3 years of it.
Basically they have given you enough information to determine what t is.
tant=4/5
therefore t must be 38.65980825...
but wait, if we look at the tan graph we also see that 4/5 or 0.8 can also occur at 38.65 plus pi or 3pi or 5pi etc.
They have also told us that cost must be less than zero, therefore if we look at the cos graph we see that cos can only be zero between 90 and 270, or 450 and 630 etc. This means that t must fall on the tan graph in those same areas as allowed on the cos chart.
So we see that t can only meet the criteria when it equals 38.65+180 or 540 etc. written out this is
t=(38.65+pi+2Xpi)
(I am giving the very general case in that equation)
So the first question a
cost=cos(38.65+pi+0)=-0.78, which meets the criteria of cost<0
second question
sin(-t) = sin -(38.65+pi)=0.624
third
cot(t+pi) = 38.65+pi+pi =1.25
fourth
sec(t+6pi)=sec(38.65+pi+6pi) or 1/cos(...)=1.28
of course you could just realize too that the 6pi is just the 2xpi term in my equation above and get rid of it as it doesnt affect the final answer. This goes for the third question too.
hope this helps.
Stocker.
If you need actual help in calculus, just ask, I did 3 years of it.
Originally posted by redrims
Fellow protege owners, I really, really, need help. I know it probably looks kinda of silly putting my math homework on the forum, but no one on my floors knows this. Its amazing because the ppl that I've asked are in higher math than me like 140 and 220. And the math websites are confusing.
Anywho, here's the problem: (For pie, I just used 3.14...../I hope y'all know what it means and not to actually use 3.14, but the sign pie)
Suppose tant=4/5 and cost<0. Find each of the following:
a. cos(-t) b. sin(-t) c. cot(t+3.14.....)
d. sec(t+6[pie])
I really hope this makes sense and thanks.
Fellow protege owners, I really, really, need help. I know it probably looks kinda of silly putting my math homework on the forum, but no one on my floors knows this. Its amazing because the ppl that I've asked are in higher math than me like 140 and 220. And the math websites are confusing.
Anywho, here's the problem: (For pie, I just used 3.14...../I hope y'all know what it means and not to actually use 3.14, but the sign pie)
Suppose tant=4/5 and cost<0. Find each of the following:
a. cos(-t) b. sin(-t) c. cot(t+3.14.....)
d. sec(t+6[pie])
I really hope this makes sense and thanks.
Last edited by stocker; November-26th-2002 at 01:10 PM.
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