# Thread: Search for an 'elegant' solution

1. ## Search for an 'elegant' solution

Given,

$x=\frac{\sin^3 p}{\cos^2 p }$

$y=\frac{\cos^3 p}{\sin^2 p }$

and $\sin p+\cos p=\frac1{2}$

Find the value of $x+y$

here is my solution:

Spoiler:

$2\sin p\cos p=\frac{-3}{4}$

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

$x+y =\frac{\sin^5+\cos^5}{\sin^2p\cos^2p}$

$\Rightarrow x+y=\frac{(\sin p+\cos p)(\sin^4p-\sin^3p\cos p+\sin^2p\cos^2p-\sin p\cos^3p+\cos^4p)}{\sin^2p\cos^2p}$

$\Rightarrow x+y=\frac{(\sin p+\cos p)^4-5\sin^3p\cos p-5\sin^2p\cos^2p-5\sin p\cos^3p}{2\sin^2p\cos^2p}$

$\Rightarrow x+y=\frac{\frac1{16}-5\sin p\cos p(\sin^2p-\sin p\cos p+\cos^2p)}{\frac{2\times9}{64}}$

$\Rightarrow x+y=\frac{\frac1{16}-5\sin p\cos p\{(\sin p+\cos p)^2-\sin p\cos p\}}{\frac{2\times9}{64}}$

$\Rightarrow x+y=\frac{\frac1{16}+\frac{5\times3}{8}(\frac1{4}+ \frac{3}{8})}{\frac{2\times9}{64}}$

$\Rightarrow x+y=\frac{\frac{79}{64}}{\frac{18}{64}}$

$\Rightarrow x+y=\frac{79}{18}$

2. Originally Posted by great_math
Given,

$x=\frac{\sin^3 p}{\cos^2 p }$

$y=\frac{\cos^3 p}{\sin^2 p }$

and $\sin p+\cos p=\frac1{2}$

Find the value of $x+y$

here is my solution:

$2\sin p\cos p=\frac{-3}{4}$

I hope you find this elegant!!!

$x=\frac{sin^3p}{cos^2p}=\frac{sinp}{cos^2p}-sinp$

simlilarly,

$y=\frac{cosp}{sin^2p}-cosp$

$x+y$

$=\frac{sinp}{cos^2p}-sinp + \frac{cosp}{sin^2p}-cosp$

$=\frac{sinp}{cos^2p}+ \frac{cosp}{sin^2p}-cosp-sinp$

$= \frac{cos^3p+sin^3p}{cos^2psin^2p}-(cosp+sinp)$

$= \frac{(cosp+sinp)^3-3cospsinp(cosp+sinp)}{(cospsinp)^2}-(cosp+sinp)$

$= \frac{\frac{1}{8}+\frac{9}{16}}{\frac{9}{64}}-\frac{1}{2}$

$=\frac{79}{18}$

3. I imagine that the use of a calculator introduces an element of inelegance into the solution, otherwise we could simply state that

$2sinpcosp=sin2p=\frac{-3}{4}$

and find the value of p and use that to solve the equation