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Trignometry

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23 Apr 2008 08:57:41 IST

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Trig Ineq

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$\text{In a right triangle ABC with} \ \angle A = 90^o \ \text{and sides a,b and c,} \\ \\ \text{Prove that} \ b^n + c^n < a^n$

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N.KANDASWAMY

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23 Apr 2008 09:01:27 IST

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assume a=5,b=4,c=3,n=3.then b^n+c^n

Anand Hegde

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23 Apr 2008 09:09:49 IST

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but sir how does your inequality hold for n=1 and n=2?

Anand Hegde

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23 Apr 2008 09:11:31 IST

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and kanda is that your proof?

prashant

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23 Apr 2008 09:31:56 IST

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but when n=2 it is a right angled pythogores theorm... not satisfying..!!

snehashis choudhury

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23 Apr 2008 09:37:01 IST

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b^n +c^n < a^n
(b/a)^n + (c/a)^n < 1
now, b/a = sin@ & c/a=cos@ [A=90]
we have to prove sin^n@ + cos^n@ < 1
since sin^n@ + cos^n@ is a decreasing function if n is the variable
and for n=2, sin^n@ + cos^n@=1
therefore for any value of n above 2, sin^n@ + cos^n@ < 1
hence proved.

prashant

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23 Apr 2008 09:57:55 IST

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but isnt it just hidden trial method.. ? consider a case in which such inequality is given but the difference there is n=100 or 200 or so... then will u put the values from 1 to n ??

sandeep ramesh

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23 Apr 2008 10:10:20 IST

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the question should state it for n >= 3 :)

The proof is a one liner in saying!

sinC = c/a, cosC = b/a implied divide both sides by a

You get sin^n C + cos^n C < = sin^2C + cos^2C = 1

QED

PS: I believe this is some olympiad sum?

EDIT: Just saw that someone had already posted the answer.

Hari Shankar

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23 Apr 2008 11:19:50 IST

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Yet again, I have forgotten to put in the condition, which some of u have already noted. n>2 shud be it. This is from some Ramaiah Coaching entrance material I retrieved which coincidentally is also posted on mathlinks. some poor sod gave a very complicated proof, so i wanted to see how our chaps will fare

Sagar Vaze

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23 Apr 2008 11:43:36 IST

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hey why so complicated proofs
take pythagoras theorem and raise it to required power

b^2+c^2=a^2
say n=3

so

(b^2+c^2)^3/2=a^3

now lhs is b^3 + c^3 + some positive numbers
so b^3+c^3<a^3
this can be said for any n

abhishek gupta

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23 Apr 2008 12:27:03 IST

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let b=asin@ , c=acos@

{ sin^n(@)+cos^n(@) }/2< {(sin@+cos@)/2}^n< {1/sqrt(2)}n

or, sin^n(@) + cos^n(@) < { 1/2}^(n-2)/2 < 1 for n>2

so a^n{ sin^n(@) + cos^n(@)}< a^n

jagdeep singh

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23 Apr 2008 12:32:43 IST

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put a=b2+c2 and then apply am gm hm ineqalities to prove it

give me rates

Hari Shankar

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23 Apr 2008 13:44:48 IST

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As our consultant

Rohit a.k.a sandeepramesh has pointed out, the only thing you have to note is that since $\sin \theta \le 1, \sin^n \theta \le \sin^2 \theta$ for $n \ge 2$ and ditto for cosine

So, the problem can be solved as $\sin^n \theta + \cos^n \theta \le \sin^2 \theta + \cos^2 \theta \le 1$

sandeep ramesh

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23 Apr 2008 15:17:12 IST

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consultant? :rot

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