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Trignometry
23 Feb 2007 16:58:57 IST
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23 Feb 2007 18:37:36 IST
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Hey joyfrancis,you've made a mistake here.How can you assume that the values of all the trigonometric functions are positive??
Only if all the values are positive<first quad> AM>=GM can be used.
I think the answer is 2. <x=135 degrees>
soln:
see, for min value,we need to have as many negative trig. functions as possible.but at least 2<sin,cosec or tan,cot or cos,sec> of the 6 trig functions will always be positive.
their least value value by AM>GM will be 2.
for the other 4,<let them be a,1/a,b,1/b> by AM>GM max value will be -4.
now,
2+a+1/a+b+1/b<=-2
for any value less than this,the sum will go below -2 ,implies mod value will increase. therefore answer is 2,but I hope someone can come up with a better solution.
That solves the problem!
I think the answer is 2. <x=135 degrees>
soln:
see, for min value,we need to have as many negative trig. functions as possible.but at least 2<sin,cosec or tan,cot or cos,sec> of the 6 trig functions will always be positive.
their least value value by AM>GM will be 2.
for the other 4,<let them be a,1/a,b,1/b> by AM>GM max value will be -4.
now,
2+a+1/a+b+1/b<=-2
for any value less than this,the sum will go below -2 ,implies mod value will increase. therefore answer is 2,but I hope someone can come up with a better solution.
That solves the problem!
24 Feb 2007 18:29:58 IST
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pheeeeeew!
took me some time to solve
took me some time to solve
here is the solution
put
x=y-3pi/4
sinx=-(cosy+siny)/sqroot(2)
cosx=-(cosy-siny)sqroot(2)
sinx+cosx=-sqroot(2)cosy
tanx+cotx=2(cos2y-sin2y)
sec x+cosecx=-sqroot(2)cosy(cos2y)
then
| sinx + cosx + tanx + cotx + secx + cosecx |
=|t+2/(1+t)| where t=sqroot(2)*cosy
now
=|t+2/(1+t)|>=2sqroot(2)-1 (using A.M >=G.M)
hope this helps
hope this helps
i have tried my best to explain this one
bye!!
cheers!
cheers!
25 Feb 2007 12:15:50 IST
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Hi Shakir!
I dont see how you are justified in using AM>=GM
as you cant be sure that the terms are all positive.even if they are,see this..
By AM-GM you get limiting case of t=2/1+t
implies t^2 + t -2=0
implies t= -2 or 1
if t=1 ,sqroot(2)*cosy =1 implies cos y=1/root2
implies y=pi/4 or 7pi/4,implies x=-pi/2 or pi respectively.
but in these cases either tan or cot will attain plus or minus infinity.therefore not defined.
t=-2 is not a possible case as cosy=-root2
note: if x=pi, | sinx + cosx + tanx + cotx + secx + cosecx |=|0 + (-1) + 0 + (-infinity) + (-1) + (+infinity)| as ((-infinity) + (+infinity)) appears,I am not sure what value it will take.
please correct me if i am wrong anywhere.
I dont see how you are justified in using AM>=GM
as you cant be sure that the terms are all positive.even if they are,see this..
By AM-GM you get limiting case of t=2/1+t
implies t^2 + t -2=0
implies t= -2 or 1
if t=1 ,sqroot(2)*cosy =1 implies cos y=1/root2
implies y=pi/4 or 7pi/4,implies x=-pi/2 or pi respectively.
but in these cases either tan or cot will attain plus or minus infinity.therefore not defined.
t=-2 is not a possible case as cosy=-root2
note: if x=pi, | sinx + cosx + tanx + cotx + secx + cosecx |=|0 + (-1) + 0 + (-infinity) + (-1) + (+infinity)| as ((-infinity) + (+infinity)) appears,I am not sure what value it will take.
please correct me if i am wrong anywhere.
25 Feb 2007 13:33:11 IST
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Hi friends
Shakir is Correct. I have got a good support for that. The graph of | sinx + cosx + tanx + cotx + secx + cosecx | is shown below.
The minimum value 2 sqrt(2) - 1 is attained at 2npi +5 pi /3 . The first value is 300 degrees .
Good work shakir .
Cheers!
26 Feb 2007 19:00:58 IST
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ya ans is 2
here is the simplest method:
convert everything to sinx & cosx,,on simplifying we get
I sinx+cosx + 2 (1+ sinx + cosx)by sin2x I
if sinx+cosx =0 then the function wd get min value
logic is that two terms are cancelling and for x= - pi by 4 sin2x=-1
so,ans=I-2I=2
shakir is absolutely right,that's a good subjective one
but i think his method might not come in mind immediately at exam time
27 Feb 2007 17:38:40 IST
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Actually i had got the answer using calculus
i had put
sin x+cosx=t
and i got the expression
as |t+2/(t-1)|
then we had 2 pts
where f(t)= t+2/(t-1)
have two pts as critical and valut of f(t) at those pts came out to
be f(t)=2sqroot(2)+1 and f(t)=-2sqroot(2)+1
and then we had to take their abslolute values
and we get the answer
and we get the answer
i had to avoid calculus approach.
to make solution understandable to everyone
newayz!!!
we could have used many substitutions!
we could have used many substitutions!
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cosx=b
tanx=c
eqn becomes a+b+c+1/a+1/b+1/c
a+1/a + b+1/b + c+1/c
(using the concept of AM >= GM , if we take two numbers x and 1/x
their arithemetic mean is - x + 1/x whole divided by 2
and geometric mean = 1
therefore x + 1/x > = 2
hence minimum value of the expression is
2+2+2
= 6
There You Go Buddy!