$\frac{{(sin x)}^{4}}{x} + \frac{{(cos x)}^{4}}{y} = \frac{1}{x + y}$ $(sin x)^4 / x + (cos x )^4 /y = 1/(x+y)$ Then prove that $\frac{{(sin x)}^{12}}{x^{5}} + \frac{{(cos x)}^{12}}{y^{5}} = \frac{1}{{(x + y)}^{5}}$ $(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5$ ?

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$\frac{{(sin x)}^{4}}{x} + \frac{{(cos x)}^{4}}{y} = \frac{1}{x + y}$ $(sin x)^4 / x + (cos x )^4 /y = 1/(x+y)$ Then prove that $\frac{{(sin x)}^{12}}{x^{5}} + \frac{{(cos x)}^{12}}{y^{5}} = \frac{1}{{(x + y)}^{5}}$ $(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5$ ?

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Answer 1 · 2016-09-18T17:14:19

See below.

Explanation:

Solving for $y$ $y$

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

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

Now, substituting back in

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

we can easily verify that the equality is observed.

Another way is making

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

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

then

$y^{5} {sin}^{12} x + y^{5} {sin}^{10} x {cos}^{2} x = {(\frac{x y}{x + y})}^{5}$ $y^5sin^12x+y^5sin^10xcos^2x=((xy)/(x+y))^5$ so

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

${sin}^{10} x = {⎛ ⎝ \frac{x}{x + x \frac{{cos}^{2} x}{{sin}^{2} x}} ⎞ ⎠}^{5}$ $sin^10x=(x/(x+xcos^2x/sin^2x))^5$ et voila!

Answer 2 · 2016-09-19T04:47:21

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

$\Rightarrow (\frac{x + y}{x}) {sin}^{4} x + (\frac{x + y}{y}) {cos}^{4} x = 1$ $=>((x+y)/x)sin^4x+((x+y)/y)cos^4x=1$

$\Rightarrow (1 + \frac{y}{x}) {sin}^{4} x + (1 + \frac{x}{y}) {cos}^{4} x = 1$ $=>(1+y/x)sin^4x+(1+x/y)cos^4x=1$

$\Rightarrow {({sin}^{2} x + {cos}^{2} x)}^{2} - 2 {sin}^{2} x {cos}^{2} x + (\frac{y}{x}) {sin}^{4} x + (\frac{x}{y}) {cos}^{4} x = 1$ $=>(sin^2x+cos^2x)^2-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=1$

$\Rightarrow 1 - 2 {sin}^{2} x {cos}^{2} x + (\frac{y}{x}) {sin}^{4} x + (\frac{x}{y}) {cos}^{4} x = 1$ $=>1-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=1$

$\Rightarrow - 2 {sin}^{2} x {cos}^{2} x + (\frac{y}{x}) {sin}^{4} x + (\frac{x}{y}) {cos}^{4} x = 0$ $=>-2sin^2xcos^2x+(y/x)sin^4x+(x/y)cos^4x=0$

$\Rightarrow {(\sqrt{\frac{y}{x}} {sin}^{2} x - \sqrt{\frac{x}{y}} {cos}^{2} x)}^{2} = 0$ $=>(sqrt(y/x)sin^2x-sqrt(x/y)cos^2x)^2=0$

$\Rightarrow \sqrt{\frac{y}{x}} {sin}^{2} x = \sqrt{\frac{x}{y}} {cos}^{2} x$ $=>sqrt(y/x)sin^2x=sqrt(x/y)cos^2x$

$\Rightarrow \frac{{sin}^{2} x}{x} = \frac{{cos}^{2} x}{y}$ $=>sin^2x/x=cos^2x/y$

By addendo

$\Rightarrow \frac{{sin}^{2} x}{x} = \frac{{cos}^{2} x}{y} = \frac{{sin}^{2} x + {cos}^{2} x}{x + y} = \frac{1}{x + y}$ $=>sin^2x/x=cos^2x/y=(sin^2x+cos^2x)/(x+y)=1/(x+y)$

$\Rightarrow {sin}^{2} x = \frac{x}{x + y}$ $=>sin^2x=x/(x+y)$

And

$\Rightarrow {cos}^{2} x = \frac{y}{x + y}$ $=>cos^2x=y/(x+y)$

Now

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

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

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

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

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

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$\frac{{(sin x)}^{4}}{x} + \frac{{(cos x)}^{4}}{y} = \frac{1}{x + y}$ $(sin x)^4 / x + (cos x )^4 /y = 1/(x+y)$ Then prove that $\frac{{(sin x)}^{12}}{x^{5}} + \frac{{(cos x)}^{12}}{y^{5}} = \frac{1}{{(x + y)}^{5}}$ $(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5$ ?

2 Answers

Explanation:

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(sinx)4x+(cosx)4y=1x+y(sin x)^4 / x + (cos x )^4 /y = 1/(x+y) Then prove that (sinx)12x5+(cosx)12y5=1(x+y)5(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5 ?

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Explanation:

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$\frac{{(sin x)}^{4}}{x} + \frac{{(cos x)}^{4}}{y} = \frac{1}{x + y}$ $(sin x)^4 / x + (cos x )^4 /y = 1/(x+y)$ Then prove that $\frac{{(sin x)}^{12}}{x^{5}} + \frac{{(cos x)}^{12}}{y^{5}} = \frac{1}{{(x + y)}^{5}}$ $(sin x)^12 /x^5 + (cos x)^12 /y^5 = 1/(x+y)^5$ ?