trigno tmh

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Discussion Response Post to: trigno tmh

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®µD®A (2710)

Blazing goIITian

total posts: 2717
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well its actually 79/18

here is a messy solution from me :D

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

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

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

[IMG]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/6/3/0638cc54ee34d87aa803d78fe3ec072aa8749bf5.gif[/IMG] :)

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