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Trignometry

Blazing goIITian

Joined:	9 Mar 2008
Post:	555

28 Jun 2009 16:53:47 IST

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347

trigno tmh

None

x=[(sinp)^3]/(cosp)^2

y=(cosp)^3/(sinp)^2

given sinp+cosp=1/2----------(1)
find x+y

my work
i square (1) and get sin2p=-3/4

then sin2p=(2tanp)/(1+(tanp)^2)
the value of tan p is coming out to be irrational so sin p and cos p is also irrationals
again stuck here .answer given is in rational form.
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Comments (4)

NugoRama

Blazing goIITian

Joined: 11 Mar 2009

Posts: 5517

28 Jun 2009 16:59:30 IST

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HARJEET GANDHI

Cool goIITian

Joined: 30 May 2009

Posts: 58

28 Jun 2009 18:31:35 IST

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check your Q

TEJASWI V SHETTY.

Hot goIITian

Joined: 15 Feb 2009

Posts: 186

29 Jun 2009 12:30:29 IST

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I GOT D ANSWER AS 79/128.

IF THIS IS D RIGHT ANSWER DEN LET ME KNOW,I'LL TRY TO GIVE D COMPLETE SOLN.

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

29 Jun 2009 12:55:32 IST

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well its actually 79/18

here is a messy solution from me :D

2sin pcos p=rac{-3}{4}

$\Rightarrow \sin p\cos p=\frac{-3}{8}$

$x+y =\frac{\sin^5+\cos^5}{\sin^2p\cos^2p} =\frac{(\sin p+\cos p)(\sin^4p-\sin^3p\cos p+\sin^2p\cos^2p-\sin p\cos^3p+\cos^4p)}{\sin^2p\cos^2p}\\\\\\\Rightarrow x+y=\frac{(\sin p+\cos p)^4-5\sin^3p\cos p-5\sin^2p\cos^2p-5\sinp\cos^3p}{2\sin^2p\cos^2p}\\\\\Rightarrow x+y=\frac{\frac1{16}-5\sin p\cos p(\sin^2p-\sin p\cos p+\cos^2p)}{\frac{2\times9}{64}}\\\\\Rightarrow x+y=\frac{\frac1{16}-5\sin p\cos p\{(\sin p+\cos p)^2-\sin p\cos p\}}{\frac{2\times9}{64}}\\\\\Rightarrow x+y=\frac{\frac1{16}+\frac{5\times3}{8}(\frac1{4}+\frac{3}{8})}{\frac{2\times9}{64}}\\\\\Rightarrow x+y=\frac{\frac{79}{64}}{\frac{18}{64}}\\\\\\\Rightarrow x+y=\frac{79}{18}$

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