(sin x)^4 / x + (cos x )^4 /y = 1/(x+y) Then prove that (sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5 ?

Sep 18, 2016

See below.

Explanation:

Solving for $y$

${\sin}^{4} \frac{x}{x} + {\cos}^{4} \frac{x}{y} = \frac{1}{x + y}$ we obtain after some simplifications

$y = x {\left(\cos \frac{x}{\sin} \left(x\right)\right)}^{2} = x {\cot}^{2} \left(x\right)$

Now, substituting back in

${\left(\sin x\right)}^{12} / {x}^{5} + {\left(\cos x\right)}^{12} / {y}^{5} = \frac{1}{x + y} ^ 5$

we can easily verify that the equality is observed.

Another way is making

${y}^{5} {\sin}^{12} x + {y}^{5} {\cos}^{12} x = {\left(\frac{x y}{x + y}\right)}^{5}$ and using the previous result

$y {\sin}^{2} x = x {\cos}^{2} x \to {y}^{5} {\sin}^{10} x = {x}^{5} {\cos}^{10} x$

then

${y}^{5} {\sin}^{12} x + {y}^{5} {\sin}^{10} x {\cos}^{2} x = {\left(\frac{x y}{x + y}\right)}^{5}$ so

${y}^{5} {\sin}^{10} x = {\left(\frac{x y}{x + y}\right)}^{5}$ so

${\sin}^{10} x = {\left(\frac{x}{x + x {\cos}^{2} \frac{x}{\sin} ^ 2 x}\right)}^{5}$ et voila!

Sep 19, 2016

GIVEN
${\sin}^{4} \frac{x}{x} + {\cos}^{4} \frac{x}{y} = \frac{1}{x + y}$

$\implies \left(\frac{x + y}{x}\right) {\sin}^{4} x + \left(\frac{x + y}{y}\right) {\cos}^{4} x = 1$

$\implies \left(1 + \frac{y}{x}\right) {\sin}^{4} x + \left(1 + \frac{x}{y}\right) {\cos}^{4} x = 1$

$\implies {\left({\sin}^{2} x + {\cos}^{2} x\right)}^{2} - 2 {\sin}^{2} x {\cos}^{2} x + \left(\frac{y}{x}\right) {\sin}^{4} x + \left(\frac{x}{y}\right) {\cos}^{4} x = 1$

$\implies 1 - 2 {\sin}^{2} x {\cos}^{2} x + \left(\frac{y}{x}\right) {\sin}^{4} x + \left(\frac{x}{y}\right) {\cos}^{4} x = 1$

$\implies - 2 {\sin}^{2} x {\cos}^{2} x + \left(\frac{y}{x}\right) {\sin}^{4} x + \left(\frac{x}{y}\right) {\cos}^{4} x = 0$

$\implies {\left(\sqrt{\frac{y}{x}} {\sin}^{2} x - \sqrt{\frac{x}{y}} {\cos}^{2} x\right)}^{2} = 0$

$\implies \sqrt{\frac{y}{x}} {\sin}^{2} x = \sqrt{\frac{x}{y}} {\cos}^{2} x$

$\implies {\sin}^{2} \frac{x}{x} = {\cos}^{2} \frac{x}{y}$

By addendo

$\implies {\sin}^{2} \frac{x}{x} = {\cos}^{2} \frac{x}{y} = \frac{{\sin}^{2} x + {\cos}^{2} x}{x + y} = \frac{1}{x + y}$

$\implies {\sin}^{2} x = \frac{x}{x + y}$

And

$\implies {\cos}^{2} x = \frac{y}{x + y}$

Now

${\sin}^{12} \frac{x}{x} ^ 5 + {\cos}^{12} \frac{x}{y} ^ 5$

$= {\left({\sin}^{2} x\right)}^{6} / {x}^{5} + {\left({\cos}^{2} x\right)}^{6} / {y}^{5}$

$= {\left(\frac{x}{x + y}\right)}^{6} / {x}^{5} + {\left(\frac{y}{x + y}\right)}^{6} / {y}^{5}$

$= \frac{x}{x + y} ^ 6 + \frac{y}{x + y} ^ 6$

$= \frac{1}{x + y} ^ 5$